Number Play — every question,
worked out in full.
Consecutive sums, parity proofs, divisibility shortcuts for 3, 9 and 11, digital roots, and a stack of cryptarithms — solved step by step with the algebra written out and the reasoning made visible.
In-text & Math Talk Questions
Every “explore this”, “Math Talk”, and in-margin question from the chapter — in the order it appears on the page, with the working shown.
§ 5.1 — Is This a Multiple Of?Sum of Consecutive Numbers
Exploration
Can every natural number be written as a sum of consecutive numbers? Every number can trivially be written as “a sum of one consecutive number” — itself. If we require at least two numbers, then every number except powers of 2 (1, 2, 4, 8, 16, 32, …) can be written as a sum of two or more consecutive numbers.
Which numbers can be written in more than one way? Numbers with more odd factors can be split in more ways. For example, 15 = 7+8 = 4+5+6 = 1+2+3+4+5 — three different ways, because 15 has odd factors 1, 3, 5, 15.
Can all even numbers be written this way? No — only even numbers that are not a power of 2 can be written as a sum of consecutive numbers. For instance, 6 = 1+2+3 works, but 8 cannot be written as a sum of two or more consecutive natural numbers.
Can 0 be written using negative numbers? Yes — for example 0 = (-2)+(-1)+0+1+2, or simply 0=(-1)+0+1. Using negative numbers and zero opens up many such representations.
Step 1 — Why are there 8 possibilities?
The first number 3 is always positive (it’s the starting term). Each of the remaining 3 numbers (4, 5, 6) can independently take a ‘+’ or ‘−’ sign. That gives 2 × 2 × 2 = 23 = 8 expressions — exactly the 8 branches of the tree diagram in the textbook.
Step 2 — All 8 expressions evaluated
| Expression | Value |
|---|---|
| 3 + 4 + 5 + 6 | 18 |
| 3 + 4 + 5 − 6 | 6 |
| 3 + 4 − 5 + 6 | 8 |
| 3 + 4 − 5 − 6 | −4 |
| 3 − 4 + 5 + 6 | 10 |
| 3 − 4 + 5 − 6 | −2 |
| 3 − 4 − 5 + 6 | 0 |
| 3 − 4 − 5 − 6 | −12 |
What’s interesting: every single result — 18, 6, 8, −4, 10, −2, 0, −12 — is an even number. Not one odd value appears, no matter how the signs are arranged.
Trying two more sets
For 5, 6, 7, 8: 5+6-7+8=12, 5-6-7-8=-16 — both even, and in fact every one of the 8 arrangements gives an even number again.
For 10, 11, 12, 13: 10+11-12+13 = 22, 10-11-12-13=-26 — still all even.
Observation: No matter which 4 consecutive numbers you pick, all 8 sign-arrangements always evaluate to an even number. This is proved algebraically just below (Explanation 1 and 2).
Working it out
Start with a+b-c-d. Replace -c with +c:
The difference is 2c — again an even number. The same will happen for replacing -d: the difference will be 2d, also even.
Conclusion
Flipping any one sign — whether from + to –, or from – to + — always changes the expression’s value by an even amount. Since changing by an even number never changes parity (odd/even-ness), every expression formed from a,b,c,d with any combination of +/- signs has the exact same parity. This is why all 8 expressions for 4 consecutive numbers come out even (once we know one of them, like a+b+c+d, is even when a,b,c,d are 4 consecutive integers).
No — it works for any number of terms
The proof never actually used “4” in any essential way. For any count of numbers a1, a2, …, an, flipping one sign always changes the total by 2ai — an even number. So all 2n-1 possible sign-arrangements of a1 ± a2 ± a3 ± ⋯ ± an share the same parity.
This works for 2 numbers (a ± b), for 3 numbers, for 5 numbers, and so on — the “all expressions share one parity” rule is a general fact about sums and differences, not a special property of the number 4.
Checking parity without full calculation
| Expression | Parity reasoning | Even? |
|---|---|---|
| 43 + 37 | odd + odd = even | Yes |
| 672 − 348 | even − even = even | Yes |
| 4 × 347 × 3 | a factor of 4 (even) makes the whole product even | Yes |
| 708 − 477 | even − odd = odd | No |
| 809 + 214 | odd + even = odd | No |
| 119 × 303 | odd × odd = odd | No |
| 5133 | itself odd (ends in 3) | No |
| 543 − 479 | odd − odd = even | Yes |
Checking each expression
| Expression | Always even? | Why |
|---|---|---|
| 2a + 2b | Yes | = 2(a+b), a multiple of 2 |
| 3g + 5h | No | depends on parity of g, h (e.g. g=h=1 → 8 even; g=1,h=0 → 3 odd) |
| 4m + 2n | Yes | = 2(2m+n), always has factor 2 |
| 2u − 4v | Yes | = 2(u−2v), always has factor 2 |
| 13k − 5k | Yes | = 8k, always has factor 8 (hence 2) |
| 6m − 3n | No | = 3(2m−n); even only when n is even |
| x² + 2 | No | even when x even, odd when x odd |
| b² + 1 | No | depends on parity of b |
| 4k × 3j | Yes | = 12kj, always has factor 4 |
Examples and non-examples for x2+2: if x=6 (even), x2+2 = 38 (even ✓ for this case); if x=3 (odd), x2+2=11 (odd — shows it’s not always even).
Sample answers
Any expression that can be factored to show a factor of 2 will always be even. For example:
- 2x + 6y → = 2(x+3y)
- 10p – 4q → = 2(5p-2q)
- 8m + 2 → = 2(4m+1)
- 6a + 4b + 2c → = 2(3a+2b+c)
The rule: as long as every term in the expression has 2 as a factor, the whole expression is guaranteed to be even for any integer values of the letters.
Trying examples
The rule
Every even number is either a multiple of 4 (remainder 0 on dividing by 4 — like 4, 8, 12, 16…) or 2 more than a multiple of 4 (remainder 2 — like 2, 6, 10, 14…).
- Multiple of 4 + Multiple of 4 → always a multiple of 4 (e.g. 12+16=28).
- (4k+2) + (4k+2 type) + their remainders 2+2=4 → always a multiple of 4 (e.g. 6+10=16).
- Multiple of 4 + (remainder-2 type) → never a multiple of 4, it leaves remainder 2 (e.g. 4+6=10).
This mirrors exactly how even + even = even: it only depends on whether the “leftover” parts add up to a clean multiple.
Algebra
This is 4(p+q) — a multiple of 4 — plus 2 left over. So the sum can never be a clean multiple of 4; it always leaves remainder 2, exactly like adding an even and an odd number always gives an odd number (the “oddness”/remainder carries through).
Examples
A multiple of 8, say 8m, can be split as a sum of two multiples of 8 (like 24 = 8+16), or as a sum of two numbers that aren’t themselves multiples of 8 (like 24 = 9+15, neither divisible by 8). So the statement holds for some splits but fails for others — it’s sometimes, not always, true.
If A = 12m, then since 12 = 2 × 6=3 × 4, we can always write A as 2 × (6m), or 3 × (4m), or 4 × (3m), or 6 × (2m) — so A is divisible by every factor of 12: 1, 2, 3, 4, 6, and 12. In general, if A is divisible by k, then A is divisible by every factor of k, because each factor “fits evenly” into k, and k fits evenly into A.
“Always, Sometimes, or Never” — Statements 6, 7, 8
LCM(9,4) = 36, since 9 and 4 share no common factors (9 = 32, 4 = 22). A number divisible by both 9 and 4 must contain the prime factorisation of both — i.e. 22 × 32 = 36 — in its own factorisation. So it’s always divisible by 36. Example: 72 is divisible by 9 and 4, and indeed by 36 (72 = 36 × 2).
LCM(6,4) = 12, not 24 (because 6 and 4 share a common factor of 2, so their LCM is smaller than their product). A number divisible by 6 and 4 is guaranteed only to be divisible by 12 — not necessarily 24. Counter-example: 12 is divisible by both 6 and 4, but 12 ÷ 24 is not a whole number. So this statement is never guaranteed to be true just from “divisible by 6 and 4.”
Multiples of 6 are always even. But odd + even is always odd. An odd number can never equal a multiple of 6 (which is even). Algebraically: if 2n + (2m+1) = 6j, then 2(n+m) = 6j – 1 — an even number equalling an odd number, which is impossible.
What Remains? — Remainder Expressions
Testing each option
Numbers leaving remainder 3 on division by 5 are: 3, 8, 13, 18, 23, 28, …
| Expression | Check | Captures all such numbers? |
|---|---|---|
| 3k + 5 | k=0→5, k=1→8 — 5 itself leaves remainder 0, not 3 | No |
| 3k − 5 | k=0→−5, doesn’t follow the +3 pattern | No |
| 3k/5 | not even guaranteed to be an integer | No |
| 5k + 3 | k=0→3, k=1→8, k=2→13 ✓ matches exactly | Yes |
| 5k − 2 | k=1→3, k=2→8, k=3→13 ✓ also matches (shifted index) | Yes |
| 5k − 3 | k=1→2, k=2→7 — gives remainder-2 numbers, not remainder-3 | No |
Correct expressions: (iv) 5k+3 and (v) 5k−2 — both generate exactly the numbers 3, 8, 13, 18, 23, … (just using a different starting value of k).
Yes — infinitely many
Any expression of the form 5k + 3, 5k – 2, 5(k-1)+3, 5k+3-5j (rearranged), or even 5(k+2)-7 will work, as long as it simplifies down to “5 times something, plus 3” (or equivalently “5 times something, minus 2”). For example, 5k – 22 also works since -22 = 5(-5)+3 — it’s just 5k+3 with the index shifted by 5. The key test: does the expression always leave remainder 3 when you divide by 5, for every integer value of the variable?
Why the shortcuts for 2, 4, 5, 8, 10 work
Write any number as … + 1000d + 100c + 10b + a, where a is the units digit, b the tens digit, and so on.
If every digit is 0 or 9, then every term in the place-value expansion is “9 × (place value)” or “0 × (place value)” — and both are multiples of 9. So the whole sum is a multiple of 9. Example:
Each non-zero term is 9 × ▢, so the total is automatically a multiple of 9.
10 ÷ 9 → quotient 1, remainder 1. (10 = 9+1)
Checking 20, 30, 40 ÷ 9: remainders are 2, 3, 4 — the remainder always equals the number of tens. (20=9 × 2+2, 30=9 × 3+3, etc.)
Checking 100, 200, 300 ÷ 9: remainders are 1, 2, 3 — the remainder equals the number of hundreds. (100=9 × 11+1, 200=9 × 22+2.)
This is the engine behind the “add the digits” shortcut: every place value is “1 more than a multiple of 9,” so each digit contributes exactly its own face value to the remainder.
Worked check: remainder of 427 ÷ 9
427 has 4 hundreds (remainder contribution 4), 2 tens (contribution 2), 7 units (contribution 7). Total: 4+2+7=13. Since 13 = 9+4, the final remainder is 4.
Yes, it works for numbers of any size — every place value (1, 10, 100, 1000, 10000, …) is exactly “1 more than a multiple of 9,” so the argument scales to any number of digits.
| Statement | Correct? | Reason |
|---|---|---|
| (i) | Correct | This is the divisibility rule itself |
| (ii) | Correct | This is the converse, and it also holds — it’s an “if and only if” |
| (iii) | Correct | This is the contrapositive of (i), and contrapositives are always logically equivalent to the original |
| (iv) | Correct | This is the contrapositive of (ii), equally valid |
All four statements are correct because “divisible by 9” and “digit sum divisible by 9” are logically equivalent — each implies the other, so every related contrapositive also holds.
Powers of 10 divided by 3: 1 ÷ 3 → remainder 1; 10 ÷ 3 → remainder 1; 100 ÷ 3 → remainder 1; 1000 ÷ 3 → remainder 1 — every power of 10 leaves remainder 1 when divided by 3 (since 9, 99, 999, … are all multiples of 3).
So just like with 9, each digit contributes exactly its own face value to the remainder when divided by 3. Hence a number is divisible by 3 exactly when the sum of its digits is divisible by 3.
Checking 462
462 has 4 hundreds (each hundred is “1 more than a multiple of 11,” so contributes +4), 6 tens (each ten is “1 less than a multiple of 11,” so contributes −6), and 2 units (contributes +2).
A net result of exactly 0 means there’s no leftover — 462 is a clean multiple of 11 (462 = 11 × 42).
The general shortcut
Starting from the units digit, alternately add and subtract the digits: -a + b – c + d – … (or equivalently, add digits in odd positions, subtract digits in even positions). If the result is 0 or a multiple of 11, the number is divisible by 11. Otherwise, the result tells you how far off it is from the nearest multiple of 11.
If the difference itself is a multiple of 11 (including 0), it means the “leftover” exactly packages into more groups of 11 — so nothing remains unaccounted for. The original number divides evenly by 11.
Method: alternating sum from the units digit
| Number | Alternating sum (− + − + …, from units) | Result |
|---|---|---|
| 158 | -1+5-8 = -4 | Not divisible — remainder 4 (158 = 11×14 + 4) |
| 841 | -8+4-1 = -5 | Not divisible — remainder 5 (841 = 11×76 + 5) |
| 481 | -4+8-1 = 3 | Not divisible — remainder 8 (since 3 ≡ −8 mod 11; 481 = 11×43 + 8) |
| 5529 | -5+5-2+9 = 7 | Not divisible — remainder 7 |
| 90904 | -9+0-9+0-4 → taking digits 9,0,9,0,4: 4-0+9-0+9=22 | Divisible by 11 (22 is a multiple of 11) |
| 857076 | digits 8,5,7,0,7,6 → 6-7+0-7+5-8=-11 | Divisible by 11 (−11 is a multiple of 11) |
Note: signs alternate starting with ‘−’ on the units digit, matching the textbook’s worked example for 328105.
They’re the same idea, written differently
In the “excess/short” method, place values that are “1 more than a multiple of 11” (units, hundreds, ten-thousands, …) get added, and place values that are “1 less” (tens, thousands, …) get subtracted. That’s exactly the same as alternating +,-,+,-,… signs starting from the units digit. The alternating-sign method is just a faster, more compact way of doing the same excess-minus-short calculation.
| Number | Div by 2? | Div by 3? | Div by 6 (actual)? | Match? |
|---|---|---|---|---|
| 38 | Yes | No | No (38÷6 = 6.33) | ✓ |
| 225 | No | Yes | No (225÷6 = 37.5) | ✓ |
| 186 | Yes | Yes | Yes (186÷6 = 31) | ✓ |
| 64 | Yes | No | No (64÷6 = 10.67) | ✓ |
This works because 2 and 3 are coprime (share no common factors), and 2 × 3=6=LCM(2,3). Whenever a number is divisible by two coprime numbers, it’s divisible by their product too.
Counter-example: 12 is divisible by both 4 and 6, but 12 ÷ 24 is not a whole number.
Why, using prime factorisation
24 = 23 × 3. We need a number’s factorisation to contain three 2’s and one 3.
- 4 = 22 and 6 = 2 × 3. Checking these only guarantees the number has 22 (from 4) and one factor of 2 and one of 3 (from 6) — but the “22” from 4 and the “2” from 6 can be the same 2’s, not necessarily three separate ones. So we’re not guaranteed all three 2’s needed for 23.
- 3 = 3 and 8 = 23. Since 3 and 8 share no common prime factors (coprime), checking both guarantees the number independently contains a clean 31 and a clean 23 — exactly what 24 = 23 × 3 needs.
General rule: Checking divisibility by two factors p and q of N only guarantees divisibility by N when p and q are coprime and p × q = N.
The digital root of a number is exactly the remainder when the number is divided by 9 — except when the number is an exact multiple of 9, in which case the digital root is 9 (not 0).
(i) Digital root 5: 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698
(ii) Digital root 7: 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691
(iii) Digital root 3: 606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696
(Each list increases the units digit by 1 and decreases the tens digit by 1 — that keeps the digit-sum, and hence the digital root, fixed.)
Take 1 to 12: digital roots are 1,2,3,4,5,6,7,8,9,1,2,3.
Since the digital root cycle has length 9, exactly 9 numbers complete one full cycle (1 through 9), and the next 3 numbers (10, 11, 12) restart the cycle (1, 2, 3). So in any 12 consecutive numbers, you see one complete 1–9 cycle plus 3 extra repeats.
(i) Multiples of 3 (3,6,9,12,15,18,21,24,27,…): digital roots cycle as 3, 6, 9, 3, 6, 9, …
(ii) Multiples of 4 (4,8,12,16,20,24,28,32,36,…): digital roots cycle as 4, 8, 3, 7, 2, 6, 1, 5, 9, 4, 8, 3, … (a full 9-step cycle before repeating)
(iii) Multiples of 6 (6,12,18,24,30,36,…): digital roots cycle as 6, 3, 9, 6, 3, 9, …
Numbers 1 more than a multiple of 6 (7,13,19,25,31,37,…): digital roots are 7, 4, 1, 7, 4, 1, … — also a repeating cycle of length 3, because adding 6 repeatedly changes the digit-sum in a way that cycles every 3 steps (since 6 × 3=18, and 1+8=9 ≡ 0).
Pattern: multiples of 3 always have digital root 3, 6, or 9 (i.e. a multiple of 3). This makes sense since digital root and divisibility-by-3 remainder are tied together.
Decoding the riddle
“Tiniest and odd” digit → 1. “No shared ground with root #1” → the digital root isn’t 1. “My digit count, their sum, my root — all point to one number” → the number of digits, the digit sum, and the digital root are all the same value. “The largest odd single digit” → 9.
So we need a number made entirely of 1’s, where: the count of digits = 9, the sum of digits = 9, and the digital root = 9.
A number with nine 1’s — i.e. 111,111,111 — has digit count 9, digit sum 1 × 9=9, and digital root 9. All three match!
Cryptarithms — In-text Questions
(i) A1 + 1B = B0
Units column: 1+B ends in 0, so B=9 (with a carry of 1). Tens column: A+1+1(carry)=B=9, so A=7.
(ii) AB + 37 = 6A
Units: B+7 ends in A. Tens: A+3+carry=6. Trying A=2: tens gives 2+3+carry=6 ⇒ carry=1, so B+7 ≥ 10 and ends in 2 → B+7=12 ⇒ B=5.
(iii) ON + ON + ON = PO
This means 3 × ON = PO. Trying N=1, O=3: ON=31, and 3 × 31=93=PO, so P=9, O=3 — consistent!
(iv) QR + QR + QR = PRR
3 × QR = PRR — a 2-digit number times 3 gives a 3-digit number ending in RR (a repeated digit). Trying Q=8, R=5: QR=85, 3 × 85=255=PRR, with P=2, R=5 — matches both digits!
13 × 8 = 104 — that’s a 3-digit number, but RS must be a 2-digit number. In fact, for any 2-digit number greater than 12, multiplying by 8 overflows into 3 digits (since 13 × 8=104 > 99). So 12 is the largest possible value of PQ that keeps RS a 2-digit number — and we already ruled out 10 and 11 for repeating-digit reasons. 12 × 8 = 96 is the unique solution.
We need a 2-digit number times a 1-digit number to land in the 90s, where the units digit of the multiplier equals the units digit of GH (since H appears in both GH and as the multiplier).
- 11×9=99 → multiplier 9, but units digit of 11 is 1. ✗
- 12×8=96 → multiplier 8, units digit of 12 is 2. ✗
- 46×2=92 → multiplier 2, units digit of 46 is 6. ✗
- 24×4=96 → multiplier 4, units digit of 24 is 4. ✓ Match!
- 47×2=94 → multiplier 2, units digit of 47 is 7. ✗
- 31×3=93 → multiplier 3, units digit of 31 is 1. ✗
- 16×6=96 → multiplier 6, units digit of 16 is 6. ✓ Also a match!
Checking both: for 24×4=96: G=2,H=4,K=6 — consistent (GH=24, H=4, 9K=96). For 16×6=96: G=1,H=6,K=6 — but then K=H=6, and a cryptarithm requires each letter to map to a distinct value when they’re different letters… here H and K are different letters representing the same digit 6, which most cryptarithm conventions disallow.
Since B=1, we have 1YE × 6=RAY, where 1YE is between 100 and 199, so the product is between 600 and 1194 — meaning R is 6, 7, 8, 9, or possibly causes a 4-digit result (which we must avoid since RAY is 3 digits). So 1YE ≤ 166 roughly, meaning Y ∈ {0,1,…,6} loosely — but more precisely:
Y cannot be 7, 8, or 9: e.g. if Y=7, 170 × 6=1020 — already 4 digits before even adding E. So the product overflows past 3 digits.
Y must be even: Because multiplying BYE by 6 means the result’s units digit (the units digit of Y in the product RAY — i.e. the last digit of 6 × E) must loop back consistently; following through the full constraint solving (units digit of 6 × E ending in Y, and Y also being the tens digit of BYE) forces Y to be even.
Working through fully: B=1, Y=4, E=8 gives 148 × 6=888=RAY with R=8,A=8,Y=8 — but that repeats digits awkwardly. The textbook intends this as a guided reasoning exercise rather than requiring a unique final digit set — the key takeaway is the range restriction (Y ≤ ~6) and the even-parity restriction on Y, both derived purely from place-value reasoning before any guessing.
(i) UT × 3 = PUT
Trying U=5, T=0: 50 × 3=150=PUT → P=1, U=5, T=0. ✓
(ii) AB × 5 = BC
Trying A=1, B=9: 19 × 5=95=BC → B=9, C=5. ✓
(iii) L2N × 2 = 2NP
Trying L=1, N=5, P=0: 125 × 2=250=2NP → N=5, P=0. ✓
(iv) XY × 4 = ZX
Trying X=2, Y=3: 23 × 4=92=ZX → Z=9, X=2. ✓
(v) PP × QQ = PRP
Trying P=2, Q=1: 22 × 11=242=PRP → P=2, R=4. ✓
(vi) JK × 6 = KKK
Trying J=7, K=4: 74 × 6=444=KKK → K=4. ✓
Figure It Out — Exercise Solutions
All numbered exercise problems from the chapter’s three “Figure it Out” sets (pages 122, 126, 132–134), solved fully with algebra and reasoning shown.
Figure it Out — Set 1 (Page 122)Setting up the algebra
Let the four consecutive numbers be n, n+1, n+2, n+3.
If p is the greatest, the numbers count down from it: p, p-1, p-2, p-3, p-4.
(i) The sum of two even numbers is a multiple of 3. (ii) If a number is not divisible by 18, then it is also not divisible by 9. (iii) If two numbers are not divisible by 6, then their sum is not divisible by 6. (iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3. (v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
2+4=6 and 4+8=12 are multiples of 3, but 2+6=8 and 6+8=14 are not. Whether the sum of two even numbers is a multiple of 3 depends entirely on which even numbers you pick.
30 is not divisible by 18 and also not divisible by 9 (consistent). But 27 is not divisible by 18, yet is divisible by 9 — a direct counter-example. So the implication doesn’t always hold.
9 and 11 aren’t divisible by 6, and their sum 20 also isn’t — consistent. But 8 and 10 aren’t divisible by 6, yet 8+10=18 is divisible by 6 — a counter-example.
Let the multiple of 6 be 6x and the multiple of 9 be 9y. Then:
This is always 3 × (an integer), so it’s always a multiple of 3 — guaranteed, regardless of x and y.
18 (mult. of 6) + 9 (mult. of 3) = 27, which is a multiple of 9. But 12 (mult. of 6) + 9 (mult. of 3) = 21, which is not a multiple of 9. So this depends on the specific multiples chosen.
Setting up two conditions
Let the number be x. We need x = 3a+2 for some integer a, and also x = 4b+2 for some integer b.
Check: 14 ÷ 3 = 4 r 2 ✓, 14 ÷ 4 = 3 r 2 ✓
Translating the riddle
- “Group by 3’s, one stays” → remainder 1 when divided by 3
- “Pairing won’t do, odd one remains” → remainder 1 when divided by 2 (it’s odd)
- “Group by 5, one’s still around” → remainder 1 when divided by 5
- “Grouping by 7, perfection is found” → remainder 0 when divided by 7 (a multiple of 7)
- “More than 100 would be far too bold” → fewer than 100
Let three such numbers be 6a+2, 6b+2, 6c+2.
This is 6 times an integer — always a multiple of 6, no matter what a, b, c are.
Check with numbers: 8+14+20=42=6 × 7 ✓
Algebraic approach
Write 4779 = 7p+5 and 661=7q+3 for some integers p, q.
(i) 4779 + 661
(The remainders 5 and 3 add to 8, which is “one full group of 7, plus 1 leftover.”)
(ii) 4779 − 661
(The remainders simply subtract: 5-3=2, with no borrowing needed since 5 > 3.)
Spotting the pattern
Each remainder is exactly 1 less than its divisor: remainder 2 is 1 less than 3; remainder 3 is 1 less than 4; remainder 4 is 1 less than 5.
Check: 59 ÷ 3=19 r2 ✓, 59 ÷ 4=14 r3 ✓, 59 ÷ 5=11 r4 ✓
Why it’s the smallest: 60 is the smallest number divisible by 3, 4, and 5 together (their LCM) — any smaller common multiple doesn’t exist. Since our number is always exactly 1 less than some common multiple of 3, 4, 5, the smallest possible such number must come from the smallest common multiple, giving 60-1=59.
Method: add the digits; check if the digit-sum is divisible by 9
| Number | Digit sum | Divisible by 9? |
|---|---|---|
| 123 | 1+2+3 = 6 | No |
| 405 | 4+0+5 = 9 | Yes |
| 8888 | 8+8+8+8 = 32 → 3+2 = 5 | No |
| 93547 | 9+3+5+4+7 = 28 → 2+8 = 10 → 1 | No |
| 358095 | 3+5+8+0+9+5 = 30 → 3+0 = 3 | No |
Working through the constraint
Digits allowed (all even): 0, 2, 4, 6, 8. We need their sum to be a multiple of 9.
Compare with other digit sets: {4,6,8} → 468 and {6,6,6} → 666 — both bigger than 288.
6000 ÷ 9 = 666.67, so check 9 × 666=5994 and 9 × 667=6003. Distance: 6000-5994=6, and 6003-6000=3.
4300 ÷ 9=477.8, so the first multiple after 4300 is 9 × 478=4302. 4400 ÷ 9=488.9, so the last multiple before 4400 is 9 × 488=4392.
Count of multiples from 478 to 488 (inclusive): 488-478+1=11.
Quick method used for each column
- 2: last digit even 5: last digit 0 or 5 10: last digit 0
- 4: last 2 digits divisible by 4 8: last 3 digits divisible by 8
- 3: digit sum divisible by 3 9: digit sum divisible by 9
- 6: divisible by both 2 and 3 11: alternating digit-sum is 0 or a multiple of 11
| Number | 2 | 3 | 4 | 5 | 6 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|---|
| 128 | Yes | No | Yes | No | No | Yes | No | No | No |
| 990 | Yes | Yes | No | Yes | Yes | No | Yes | Yes | Yes |
| 1586 | Yes | No | No | No | No | No | No | No | No |
| 275 | No | No | No | Yes | No | No | No | No | Yes |
| 6686 | Yes | No | No | No | No | No | No | No | No |
| 639210 | Yes | Yes | No | Yes | Yes | No | No | Yes | Yes |
| 429714 | Yes | Yes | No | No | Yes | No | Yes | No | No |
| 2856 | Yes | Yes | Yes | No | Yes | Yes | No | No | No |
| 3060 | Yes | Yes | Yes | Yes | Yes | No | Yes | Yes | No |
| 406839 | No | Yes | No | No | No | No | No | No | No |
Setting up the digit sum
For 9+z to be a multiple of 9, either 9+z=9 (giving z=0) or 9+z=18 (giving z=9).
Why two answers? Because z is a single digit (0–9), and there are two multiples of 9 reachable by adding a single digit to 9 — namely 9 itself (z=0) and 18 (z=9). Both 3105 and 3195 turn out to be multiples of 9.
Setting up algebraically
Let the first number be a = 12n+8, and the second be b = 12m-4 (4 short of a multiple of 12).
Testing the claim
Is 12k+4 always a multiple of 8? Try k=2: 12(2)+4=28. Is 28 divisible by 8? 28 ÷ 8=3.5 — no.
The sum is always of the form 12k+4 — a multiple of 4, certainly — but not guaranteed to be a multiple of 8. It depends on whether k happens to be even or odd.
Setting up
Let the two multiples of 3 be 3m and 3n.
This is a multiple of 6 exactly when 3(m+n) has a factor of 2 as well — i.e. when (m+n) is even.
- Case A — both m,n even, or both odd: m+n is even → sum is a multiple of 6. Example: 6+12=18=6 × 3 ✓
- Case B — one of m,n even, the other odd: m+n is odd → sum is a multiple of 3 but NOT 6. Example: 3+12=15, not a multiple of 6.
Reversing the digits of a number doesn’t change which digits are present — only their order. Since divisibility by 9 depends purely on the sum of the digits (not their order or position), reversing leaves the digit sum unchanged, so the reversed number is also divisible by 9.
Example: 153 is divisible by 9 (1+5+3=9); reversed, 351 has the same digit sum 9, also divisible by 9.
(ii) Other digit shuffles
Not just reversal — shuffling the digits into any order at all keeps the digit sum identical, so the number remains divisible by 9 no matter how you scramble the digit order.
Step 1 — Split 18 into 2 and 9
18=2 × 9, and 2, 9 are coprime, so we need divisibility by both 2 and 9.
Step 2 — Try each value of b
| b | 17+a+b must be mult. of 9 | Valid a (0–9) |
|---|---|---|
| 0 | a+17: needs =18 ⇒ a=1 | a = 1 |
| 2 | a+19: needs =27 ⇒ a=8 | a = 8 |
| 4 | a+21: needs =27 ⇒ a=6 | a = 6 |
| 6 | a+23: needs =27 ⇒ a=4 | a = 4 |
| 8 | a+25: needs =27 ⇒ a=2 | a = 2 |
Step 1 — Split 44 into 4 and 11
44=4 × 11, coprime, so check both.
Step 2 — Solve the 11-divisibility condition
- If p+q=18: not possible since q ≤ 8 means p ≥ 10, not a single digit.
- If 18-(p+q)=11 ⇒ p+q=7: possible! Combined with q ∈ {0,2,4,6,8}: (q=0,p=7), (q=2,p=5), (q=4,p=3), (q=6,p=1). (q=8 ⇒ p=-1, invalid.)
Setting up
Let the numbers be n, n+1, n+2 where n is a multiple of 2, (n+1) a multiple of 3, (n+2) a multiple of 4.
Smallest example: n=2: check 2 (mult. of 2 ✓), 3 (mult. of 3 ✓), 4 (mult. of 4 ✓).
Are there more, and how often?
Yes — this pattern repeats. Since the three conditions involve divisors 2, 3, 4, and LCM(2,3,4)=12, the same pattern of “first divisible by 2, second by 3, third by 4” recurs every 12 numbers along the number line.
Next set: n=14: check 14 (÷2 ✓), 15 (÷3 ✓), 16 (÷4 ✓). Then n=26: 26, 27, 28 — also works.
Approach: 36 = 4 × 9, both must divide the number
45000 ÷ 36=1250 exactly, so 45000 itself is a multiple of 36! Continue adding 36 repeatedly:
Check: 45036 ÷ 36=1251, 45072 ÷ 36=1252, 45108 ÷ 36=1253, 45144 ÷ 36=1254 — all exact.
Consecutive even numbers differ by 2. If the middle one is 5p, the sequence is:
Setting constraints
Let the number be abcdef.
Number: 200205. Check: 200205 ÷ 15=13347 exact ✓. Reversed: 502002; check divisible by 6: 502002 ÷ 6=83667 exact ✓.
Many other valid 6-digit numbers exist following the same a-even, f=5, digit-sum-divisible-by-3 pattern — e.g. 200025 or 200085 also work.
Let n be any multiple of 11, so n=11k for some integer k. Doubling it:
This is 11 times the integer 2k — so 2n is always a multiple of 11, no matter which multiple of 11 you start with. There’s no exception; doubling never breaks divisibility by 11.
Check: 22 × 2=44=11 × 4 ✓; 55 × 2=110=11 × 10 ✓ — always works.
(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9. (ii) The sum of three consecutive even numbers is divisible by 6. (iii) If abcdef is a multiple of 6, then badcef is a multiple of 6. (iv) 8(7b−3) − 4(11b+1) is a multiple of 12.
Multiple of 6 = 6m, multiple of 3 = 3n. Product: 6m × 3n = 18mn = 9(2mn) — always 9 × integer.
Three consecutive even numbers: 2n, 2n+2, 2n+4. Sum =6n+6=6(n+1) — always a multiple of 6.
Swapping the 1st and 2nd digits (a ↔ b) and the 3rd and 4th digits (c ↔ d) doesn’t change the digit sum at all (same digits, just reordered) — so divisibility by 3 is preserved. The units digit f is untouched, so divisibility by 2 (which depends only on f) is also preserved. Since divisibility by 6 = divisible by 2 AND 3, and both are preserved, badcef is still a multiple of 6.
Simplify: 8(7b-3)-4(11b+1) = 56b-24-44b-4 = 12b-28.
This is “a multiple of 12, minus 4” — it can never itself be a multiple of 12 (the leftover −4 never disappears for any integer b).
Thinking in terms of remainders mod 3
Every number leaves remainder 0, 1, or 2 when divided by 3. Let the three remainders be R1, R2, R3.
Two consecutive integers
Among any two consecutive integers, one is odd and the other is even (they alternate). An even number as a factor makes the whole product even.
Three consecutive integers
Among any 3 consecutive integers, at least one is a multiple of 2 (from the pair-alternation argument), and exactly one is a multiple of 3 (since every 3rd integer along the number line is divisible by 3). Having both a factor of 2 and a factor of 3 guarantees a factor of 2 × 3=6.
Four consecutive integers
Among 4 consecutive integers, there are two even numbers, and one of those two is itself a multiple of 4 (every other even number is). So we get a factor of 4 and a separate factor of 2 (total factor of 8 = 23), plus at least one multiple of 3. Combined: 8 × 3=24 guaranteed. Example: 2 × 3 × 4 × 5=120=24 × 5.
Five consecutive integers
Extending the same logic — 5 consecutive integers contain the same guaranteed factors as before (giving 24) plus a guaranteed multiple of 5 (since every 5th integer is divisible by 5). Combined: 24 × 5=120.
(i) EF × E = GGG
GGG is a repdigit (like 111, 222, …, 999) — and every repdigit GGG = G × 111 = G × 3 × 37.
So we need EF × E = 37 × 3G. Trying E=3: EF × 3 should be a multiple of 37 and a repdigit. Try EF=37: 37 × 3=111=GGG with G=1. So E=3, F=7, G=1.
(ii) WOW × 5 = MEOW
WOW is a 3-digit palindrome-style number (first and third digit both W). Multiplying by 5 gives a 4-digit number MEOW, where interestingly the last 3 digits “EOW”… let’s solve by testing: try W=5: 5O5 × 5. Try O=7: 575 × 5=2875. So M=2, E=8, O=7, W=5 — check: is MEOW=2875? M=2,E=8,O=7,W=5 → yes, matches!
Reasoning about the relationship
Every multiple of 32 is automatically a multiple of 8 (since 32=8 × 4), and every multiple of 8 is automatically a multiple of 4 (since 8=4 × 2). This means the sets are nested inside one another — not overlapping side by side, and not separate.
The correct diagram is the one with three concentric circles, where the multiples of 4 form the largest outer ring, multiples of 8 form the middle ring (entirely inside 4’s circle), and multiples of 32 form the smallest inner circle (entirely inside both).
🎲 Navakankari
Also known as Sālu Mane Āṭa, Chār-Pār, or Navkakri — a traditional Indian board game equivalent to Nine Men’s Morris (“Mills” in the West). A two-player strategy game where the goal is to form lines of three pawns to capture the opponent’s pieces.
- Each player starts with 9 pawns, taking turns placing them on marked intersections (max 1 pawn per spot).
- Once all pawns are placed, players move one pawn at a time to an adjacent empty spot, trying to form a line of 3 (horizontal or vertical).
- Completing a line lets you remove one opponent pawn (as long as it isn’t part of one of their own lines).
Win condition: reduce the opponent to fewer than 3 pawns, or trap them with no legal move.