Chapter 4 — Expressions Using Letter-Numbers
Complete step-wise solutions for every in-text question and every “Figure it Out” exercise question — letter-numbers, algebraic expressions, simplification, like/unlike terms, and pattern-based formulas.
In‑Text Questions
The “?” marked discussion questions found inside every sectionExample 1: Shabnam is 3 years older than Aftab. Now Aftab’s age is 18 years — what will Shabnam’s age be?
Shabnam’s age \(=\) Aftab’s age \(+3 = 18+3\)
Given Aftab’s age, how will you find out Shabnam’s age? Can we write this as an expression?
We add 3 to Aftab’s age to get Shabnam’s age: Shabnam’s age = Aftab’s age + 3. Using \(a\) for Aftab’s age and \(s\) for Shabnam’s age, the algebraic expression is:
If \(a\) is 23 (Aftab’s age in years), then what is Shabnam’s age?
Replacing \(a=23\) in \(s=a+3\): \(s=23+3\)
Given the age of Shabnam, write an expression to find Aftab’s age. Use this expression to find Aftab’s age if Shabnam’s age is 20.
Since Aftab is 3 years younger than Shabnam: Aftab’s age = Shabnam’s age − 3, i.e. \(a = s-3\).
If \(s=20\): \(a=20-3\)
Example 2: Parthiv makes matchstick “L” patterns — each L uses 2 matchsticks. Find an algebraic expression to get the number of matchsticks for \(n\) L’s.
Number of matchsticks \(=2\times\)Number of L’s. Using \(n\) for the number of L’s:
Example 3: Coconut = ₹35, jaggery = ₹60/kg. How much should Ketaki pay for (i) 8 coconuts and 9 kg jaggery? Also write the general algebraic expression, and use it for 7 coconuts and 4 kg jaggery.
General expression (\(c\)=no. of coconuts, \(j\)=kg of jaggery):
- For 8 coconuts, 9 kg jaggery: \(8\times35+9\times60=280+540=820\) → ₹820
- For 7 coconuts, 4 kg jaggery: \(7\times35+4\times60=245+240=485\) → ₹485
Example 4: The perimeter of a square is \(4\times q\), where \(q\) is the sidelength. What is the perimeter of a square with sidelength 7 cm?
Perimeter \(=4\times q = 4\times 7\)
Find the values of the following expressions (using swapping, grouping, and bracket rules): 1. \(23-10\times2\) 2. \(83+28-13+32\) 3. \(34-14+20\) 4. \(42+15-(8-7)\) 5. \(68-(18+13)\) 6. \(7\times4+9\times6\) 7. \(20+8\times(16-6)\)
| 1. \(23-10\times2 = 23-20\) | 3 |
| 2. \(83+28-13+32\) | 130 |
| 3. \(34-14+20\) | 40 |
| 4. \(42+15-(8-7)=42+15-1\) | 56 |
| 5. \(68-(18+13)=68-31\) | 37 |
| 6. \(7\times4+9\times6=28+54\) | 82 |
| 7. \(20+8\times(16-6)=20+80\) | 100 |
Find an algebraic expression for the \(n\)th term of the sequence 4, 8, 12, 16, 20, 24, 28, …
These are multiples of 4, so the \(n\)th term is \(4\times n\), written as:
Check each simplification. Identify mistakes (if any), explain, and correct them.
| # | Given | Mistake? | Correct value |
|---|---|---|---|
| 1 | \(a=-4,\ 10-a=6\) | Yes — sign error | \(10-(-4)=\mathbf{14}\) |
| 2 | \(d=6,\ 3d=36\) | Yes | \(3\times6=\mathbf{18}\) |
| 3 | \(s=7,\ 3s-2=15\) | Yes | \(3\times7-2=\mathbf{19}\) |
| 4 | \(r=8,\ 2r+1=29\) | Yes | \(2\times8+1=\mathbf{17}\) |
| 5 | \(j=5,\ 2j=10\) | No mistake | \(2\times5=10\) ✔ |
| 6 | \(m=-6,\ 3(m+1)=19\) | Yes | \(3(-6+1)=3(-5)=\mathbf{-15}\) |
| 7 | \(f=3,g=1,\ 2f-2g=2\) | Yes | \(6-2=\mathbf{4}\) |
| 8 | \(t=4,b=3,\ 2t+b=24\) | Yes — multiplied \(2t+b\) instead of adding | \(8+3=\mathbf{11}\) |
| 9 | \(h=5,n=6,\ h-(3-n)=4\) | Yes — bracket sign error | \(5-(3-6)=5+3=\mathbf{8}\) |
Most errors happen from forgetting to substitute negative values carefully, or mishandling the bracket/order of operations.
Pencils sold: Day1=5, Day2=3, Day3=10 (price \(c\) each). Money earned Day1 = \(5c\). What about Day2 and Day3? Simplify \(5c+3c+10c\).
Day 2 earnings = \(3c\); Day 3 earnings = \(10c\).
By the distributive property: \(5c+3c+10c=(5+3+10)\times c\)
If \(c=₹50\), find the total amount earned by the sale of pencils. Also, write and simplify the expression for erasers sold (Day1=4, Day2=6, Day3=1, price \(d\) each). Can \(18c+11d\) be simplified further?
\(18c = 18\times50\)
Eraser earnings: \(4d+6d+1d = 11d\)
18c + 11d cannot be simplified further, because \(c\) and \(d\) are different (unlike) letter-numbers — it is already in simplest form.
Check that \(5c+3c+10c\) and \(18c\) take the same value when \(c\) is replaced by different numbers.
Let \(c=2\): \(5(2)+3(2)+10(2)=10+6+20=36\); and \(18(2)=36\). ✔ Equal.
Let \(c=10\): \(5(10)+3(10)+10(10)=50+30+100=180\); and \(18(10)=180\). ✔ Equal — confirms the two expressions are always equal.
A big rectangle (height \(n\)) is split into two smaller rectangles of widths 12 and 4, forming rectangle AEFD as the region of width \((12-4)\). Write an expression for the area of rectangle AEFD.
Shaded region AEFD, width = (12 − 4), height = n
Method (i) directly: side lengths \(n\) and \((12-4)\) → \(n\times8=8n\)
Method (ii) by subtracting: \(12n-4n\)
Chair rent = ₹40, returned = ₹6. Table rent = ₹75, returned = ₹10. Describe the procedure to find the total amount paid for \(x\) chairs and \(y\) tables, and simplify \((40x+75y)-(6x+10y)\). Could the initial expression have been written as \((40x+75y)+(-6x-10y)\)?
Procedure: Total paid at start = (rent per chair × no. of chairs) + (rent per table × no. of tables) = \(40x+75y\). Amount returned = \(6x+10y\). Net amount paid = (amount paid at start) − (amount returned).
- \((40x+75y)-(6x+10y) = 40x+75y-6x-10y\)
- Group like terms: \((40-6)x+(75-10)y\)
- \(=34x+65y\)
Yes — subtracting a bracket is the same as adding its negative, so \((40x+75y)-(6x+10y)=(40x+75y)+(-6x-10y)\) is a valid, equal way to write the same expression.
Charu’s quiz scores over 3 rounds: \(7p-3q,\ 8p-4q,\ 6p-2q\) (\(p\)=score for correct, \(q\)=penalty for wrong). What do these expressions mean? Find her scores in rounds 2 and 3 if \(p=4,q=1\). What if there’s no penalty? Find her final score after 3 rounds.
Each expression means: (correct answers × score per correct) minus (wrong answers × penalty). E.g. \(7p-3q\) = score from 7 correct answers minus penalty for 3 wrong answers.
Round 2: \(8(4)-4(1)=32-4=\mathbf{28}\)
Round 3: \(6(4)-2(1)=24-2=\mathbf{22}\)
If there’s no penalty, \(q=0\).
Final score \(=(7p-3q)+(8p-4q)+(6p-2q)=(7+8+6)p-(3+4+2)q\)
Krishita’s total score after 3 rounds is \(23p-7q\). Give some possible scores for Krishita’s individual rounds. Can we say who scored more, and by how much? Simplify \(23p-7q-(21p-9q)\).
Sample individual scores that add to \(23p-7q\): \(9p-2q,\ 7p-3q,\ 7p-2q\) (since \(9+7+7=23\) and \(2+3+2=7\)).
Difference: \(23p-7q-(21p-9q)=23p-7q-21p+9q=(23-21)p+(-7+9)q\)
Since \(p\) (score per correct answer) and \(q\) (penalty) are both positive quantities, \(2p+2q>0\), so Krishita scored more than Charu.
Simplify the expression \(4(x+y)-y\).
- \(4(x+y)-y = 4x+4y-y\)
- \(=4x+(4-1)y\)
Are the expressions \(5u\) and \(5+u\) equal? Fill the blanks by replacing \(u\) with different numbers and compare.
| u | 5u | 5+u |
|---|---|---|
| 11 | 55 | 16 |
| 2 | 10 | 7 |
| 8 | 40 | 13 |
| 5 | 25 | 10 |
Are the expressions \(10y-3\) and \(10(y-3)\) equal? Fill the diagrams for different values of \(y\) and compare.
| y | 10y − 3 | 10(y − 3) |
|---|---|---|
| 2 | 17 | −10 |
| 0 | −3 | −30 |
| 10 | 97 | 70 |
| 7 | 67 | 40 |
Look at all the corrected simplest forms from the “Mind the Mistake” table (Page 94). Is there any relation between the number of terms and the number of letter-numbers in these expressions?
Find out the formula of the number machine, given: (5,2)→8, (8,1)→15, (9,11)→7, (10,10)→10.
The rule is “two times the first number minus the second number”:
Find the formulas of the two number machines below and write the expression for each set of inputs.
Machine (i) — inputs (5,2)→5, (8,1)→7, (9,11)→18, (10,10)→18, (a,b)→?
Machine (ii) — inputs (4,1)→5, (6,0)→1, (3,2)→7, (10,3)→?, (a,b)→?
Machine (i): formula is “sum of the two numbers minus 2”:
Machine (ii): formula is “product of the two numbers plus 1”:
Now make a formula of your own, and write a few number machines as examples using it. Challenge your classmates!
Sample formula: “3 times the first number plus the second number”, i.e. \(3a+b\). Example inputs: (2,4)→10, (5,1)→16, (0,7)→7 — try to guess the rule from these!
A saree border repeats designs A, B, C, D, E, F. Design C appears at positions 3, 6, 9,… Find the formula for the \(n\)th occurrence of Design C, and similarly for Designs B and A.
Design C occurs at multiples of 3 → \(n\)th occurrence at position 3n.
Design B occurs one less than Design C’s positions (2, 5, 8, 11,…) → \(n\)th occurrence at position 3n − 1.
Design A occurs two less than Design C’s positions (1, 4, 7, 10,…) → \(n\)th occurrence at position 3n − 2.
Given a position number, can we find the design that appears there? Which design appears at Position 122? Can the remainder on dividing by 3 be used for this? Find designs at positions 99, 122, 148.
Yes — the remainder when the position is divided by 3 tells us the design:
| Remainder on ÷3 | Design |
|---|---|
| 0 (exact multiple of 3) | C |
| 1 | A |
| 2 | B |
| Position | Quotient | Remainder | Design |
|---|---|---|---|
| 99 | 33 | 0 | C |
| 122 | 40 | 2 | B |
| 148 | 49 | 1 | A |
In a 2×2 square from an endless calendar grid, the diagonal sums (12+20 and 13+19) are equal. Will this always happen in every 2×2 square? Given the top-left number is \(a\), find the other three numbers, and verify the diagonal-sum expression.
Generic 2×2 calendar square with top-left value a
Since each row moves 1 higher and each new row is 7 more (weekly cycle): number to the right of \(a\) is \(a+1\); below \(a\) is \(a+7\); diagonal to \(a\) is \(a+8\).
- Diagonal 1: \(a+(a+8)=2a+8\)
- Diagonal 2: \((a+1)+(a+7)=2a+8\)
Verification example: if \(a=15\): numbers are 15,16,22,23. Diagonals: \(15+23=38\) and \(16+22=38\); formula gives \(2(15)+8=38\) ✔.
For the plus/cross-shaped set of calendar numbers (8, 14-15-16, 22), find the sum of all 5 numbers and compare it with the centre number 15. Will this always happen? How do you show it? Find other shapes where the sum is a multiple of one number.
Plus-shaped set of 5 calendar numbers, centre = a
Sum \(=8+14+15+16+22=75\), and \(75=5\times15\) — the sum is 5 times the centre number.
Taking the centre as \(a\): top \(=a-7\), left \(=a-1\), right \(=a+1\), bottom \(=a+7\). Sum:
Another shape: any 3 numbers in a row \((n-1,\ n,\ n+1)\) always sum to \(3n\) — always a multiple (3×) of the middle number.
A pattern of joined triangles uses 3, 5, 7, 9, 11, 13,… matchsticks at Steps 1, 2, 3, 4, 5, 6. How many matchsticks are needed for Step 33, Step 84, and Step 108?
Number of matchsticks increases by 2 each step. Two equivalent formulas: \(3+2\times(y-1)\) or simplified, \(2y+1\), for step \(y\).
- Step 33: \(2(33)+1=\mathbf{67}\)
- Step 84: \(2(84)+1=\mathbf{169}\)
- Step 108: \(2(108)+1=\mathbf{217}\)
Matchsticks are placed in two orientations — horizontal (top & bottom) and diagonal (middle). In Step 2, there are 2 horizontal and 3 diagonal sticks. What are these numbers in Step 3 and Step 4? Write expressions for Step \(y\) in each orientation — do they add up to \(2y+1\)?
| Step | Horizontal | Diagonal |
|---|---|---|
| 2 | 2 | 3 |
| 3 | 3 | 4 |
| 4 | 4 | 5 |
At step \(y\): Horizontal sticks \(=y\); Diagonal sticks \(=y+1\).
Exercise Questions — “Figure it Out”
All formally numbered practice questions from the chapterWrite formulas for the perimeter of: (a) triangle with all sides equal (b) a regular pentagon (c) a regular hexagon.
Let \(a\) = length of each equal side.
| (a) Equilateral triangle | 3a |
| (b) Regular pentagon | 5a |
| (c) Regular hexagon | 6a |
Munirathna has a 20 m pipe and joins another pipe of length \(k\) metres. Give the expression for the combined length.
Complete the table for total amount Krithika has with given numbers of ₹100, ₹20 and ₹5 notes.
| ₹100 notes | ₹20 notes | ₹5 notes | Expression & total |
|---|---|---|---|
| 3 | 5 | 6 | \(3\times100+5\times20+6\times5=\mathbf{430}\) |
| 6 | 4 | 3 | \(6\times100+4\times20+3\times5=\mathbf{695}\) |
| 8 | 4 | z | \(8\times100+4\times20+z\times5=\mathbf{880+5z}\) |
| x | y | z | \(100x+20y+5z\) |
Venkatalakshmi’s roller mill takes 10 seconds to start, then 8 seconds per kg to grind. Which expression describes the time to grind \(y\) kg?
Total time = start-up time + (grinding time per kg × kg of grain) = \(10 + 8\times y\)
Write algebraic expressions (letters of your choice): (a) 5 more than a number (b) 4 less than a number (c) 2 less than 13 times a number (d) 13 less than 2 times a number.
Let the number be \(d\).
| (a) 5 more than a number | d + 5 |
| (b) 4 less than a number | d − 4 |
| (c) 2 less than 13 times a number | 13d − 2 |
| (d) 13 less than 2 times a number | 2d − 13 |
Describe real-life situations for: (a) \(8\times x+3\times y\) (b) \(15\times j-2\times k\)
(a) A shopkeeper sells a pen for ₹x and a notebook for ₹y. Abha buys 8 pens and 3 notebooks. Total cost = \(8x+3y\).
(b) A factory makes 15 chairs a day, working for \(j\) days (\(15j\) chairs made), but 2 chairs break each day for \(k\) days. Number of good chairs remaining = \(15j-2k\).
In a calendar month’s 2×3 date grid, the bottom-middle cell is \(w\). Write expressions for the other 5 cells.
| w−8 | w−7 | w−6 |
| w−1 | w | w+1 |
Add the numbers in each picture below (unknowns shown as letter-numbers). Write and simplify the corresponding expressions.
Grid 1 — \(5y,\ -6,\ x\) (top row); \(x,\ 2,\ 5y\) (bottom row):
Grid 2 — pairs of \(2p,3q,-2,3\) repeated in 4 rows:
Grid 3 — border of \(-5g\) with interior \(5k\)’s (4×4 grid):
Simplify each of the following expressions: (a) \(p+p+p+p\), \(p+p+p+q\) (b) \(p+q+p-q\), \(p-q+p-q\) (c) \(p+q-(p+q)\), \(p-q-p-q\) (d) \(2d-d-d-d\), \(2d-d-d-c\) (e) \(2d-d-(d-c)\), \(2d-(d-d)-c\) (f) \(2d-d-c-c\)
| \(p+p+p+p\) | 4p |
| \(p+p+p+q\) | 3p + q |
| \(p+q+p-q\) | 2p |
| \(p-q+p-q\) | 2p − 2q |
| \(p+q-(p+q)\) | 0 |
| \(p-q-p-q\) | −2q |
| \(2d-d-d-d\) | −d |
| \(2d-d-d-c\) | −c |
| \(2d-d-(d-c)\) | c |
| \(2d-(d-d)-c\) | 2d − c |
| \(2d-d-c-c\) | d − 2c |
Find and correct the mistakes in each simplification below.
| # | Expression | Given (wrong) | Correct simplest form |
|---|---|---|---|
| 1 | 3a + 2b | 5 | 3a + 2b (already simplest — unlike terms can’t combine) |
| 2 | 3b − 2b − b | 0 | 0 (correct!) |
| 3 | 6(p + 2) | 6p + 8 | 6p + 12 |
| 4 | (4x+3y) − (3x+4y) | x + 5 | x − y |
| 5 | 5 − (2 − 6z) | 3 − 6z | 3 + 6z |
| 6 | 2 + (x + 3) | 2x − 6 | x + 5 |
| 7 | 2y + (3y − 6) | −y + 6 | 5y − 6 |
| 8 | 7p − p + 5q − 2q | 7p + 3q | 6p + 3q |
| 9 | 5(2w+3x+4w) | 10w+15x+20w | 30w + 15x |
| 10 | 3j+6k+9h+12 | 3(j+2k+3h+4) | 3j+6k+9h+12 (already correct!) |
| 11 | 4(2r+3s+5) | −20−8r−12s | 8r + 12s + 20 |
Jowar roti = ₹30/plate, Pulao = ₹20/plate. If \(x\) plates of roti and \(y\) plates of pulao are ordered, which expression describes the total amount earned?
\(p\) customers bought only champak, \(q\) only marigold, \(r\) bought both. Every customer got one flag. How many flags were given away?
Total number of customers (regardless of what they bought) = \(p+q+r\); each gets exactly 1 flag.
A snail climbs \(u\) cm by day and slips \(d\) cm by night, for 10 days and 10 nights. (a) Expression for distance from starting position. (b) What if \(d>u\)?
(a) Net movement per day-night cycle = \((u-d)\) cm; over 10 cycles:
(b) If \(d>u\), the snail slips down more than it climbs each cycle — it moves net downward and will never reach the top of the well.
Radha cycles 5 km/day in week 1, increasing the daily distance by \(z\) km every week. How many km would she have cycled after 3 weeks?
- Week 1: daily distance = 5 km → weekly total = \(7\times5=35\) km
- Week 2: daily distance = \((5+z)\) km → weekly total = \(7(5+z)=35+7z\) km
- Week 3: daily distance = \((5+2z)\) km → weekly total = \(7(5+2z)=35+14z\) km
- Total = \(35+(35+7z)+(35+14z) = 105+21z\)
The expression \(w+2\) becomes \(4w+20\) along one path (×3, −5, +3, …×4). Fill in the missing blanks on the remaining paths.
Top-left path: \(w+2 \xrightarrow{\times3} 3w+6 \xrightarrow{-5} 3w+1\)… following the book’s actual box order (×3 then arrives at \(w-3\) oval, meaning the box before is applied to reach that label) — matching the answer key:
| Top-left path result | 3w − 9 |
| Bottom-left path (−4, −8) | w − 10 (via w − 6) |
| Bottom-right path (−4, ×3) | w − 2 → 3w − 6 |
A train stops at 3 equally-spaced stations; travel time between stations is \(t\) minutes; it stops 2 minutes at each station. (a) If \(t=4\), total travel time? (b) General algebraic expression?
4 travel-segments of t minutes each + 3 stops of 2 minutes each
There are 4 travel segments (Yahapur→Stn1→Stn2→Stn3→Vahapur) of \(t\) minutes each, and 3 stops of 2 minutes each.
(a) If \(t=4\): \(4(4)+3(2)=16+6=22\)
Simplify: (a) \(3a+9b-6+8a-4b-7a+16\) (b) \(3(3a-3b)-8a-4b-16\) (c) \(2(2x-3)+8x+12\) (d) \(8x-(2x-3)+12\) (e) \(8h-(5+7h)+9\) (f) \(23+4(6m-3n)-8n-3m-18\)
| (a) | \(3a+9b-6+8a-4b-7a+16 = (3+8-7)a+(9-4)b+(16-6)\) | 4a + 5b + 10 |
| (b) | \(9a-9b-8a-4b-16 = a-13b-16\) | a − 13b − 16 |
| (c) | \(4x-6+8x+12 = 12x+6\) | 12x + 6 |
| (d) | \(8x-2x+3+12 = 6x+15\) | 6x + 15 |
| (e) | \(8h-5-7h+9 = h+4\) | h + 4 |
| (f) | \(23+24m-12n-8n-3m-18 = 21m-20n+5\) | 21m − 20n + 5 |
Add the expressions: (a) \(4d-7c+9\) and \(8c-11+9d\) (b) \(-6f+19-8s\) and \(-23+13f+12s\) (c) \(8d-14c+9\) and \(16c-(11+9d)\) (d) \(6f-20+8s\) and \(23-13f-12s\) (e) \(13m-12n\) and \(12n-13m\) (f) \(-26m+24n\) and \(26m-24n\)
| (a) | 13d + c − 2 |
| (b) | 7f + 4s − 4 |
| (c) | 2c − d − 2 |
| (d) | −7f − 4s + 3 |
| (e) | 0 |
| (f) | 0 |
Subtract: (a) \(9a-6b+14\) from \(6a+9b-18\) (b) \(-15x+13-9y\) from \(7y-10+3x\) (c) \(17g+9-7h\) from \(11-10g+3h\) (d) \(9a-6b+14\) from \(6a-(9b+18)\) (e) \(10x+2+10y\) from \(-3y+8-3x\) (f) \(8g+4h-10\) from \(7h-8g+20\)
| (a) | −3a + 15b − 32 |
| (b) | 16y + 18x − 23 |
| (c) | 10h − 27g + 2 |
| (d) | −(3a + 3b + 32) |
| (e) | −13y − 13x + 6 |
| (f) | 3h − 16g + 30 |
Describe situations corresponding to: (a) \(8x+3y\) (b) \(15x-2x\)
(a) A fruit seller sells mangoes at ₹8 each and bananas at ₹3 each. If a customer buys \(x\) mangoes and \(y\) bananas, the total cost is \(8x+3y\).
(b) A shopkeeper has 15 pencils in a packet, priced at ₹x each. 2 pencils from the packet remain unsold. \(15x-2x\) gives the amount received for the pencils actually sold.
A straight rope cut once gives 2 pieces; folded once and cut gives 3 pieces. Find the number of pieces if folded 10 times and cut. What is the expression for \(r\) folds?
Pattern: 0 folds → 2 pieces; 1 fold → 3 pieces; each extra fold adds 1 more piece.
10 folds: \(10+2=12\)
A matchstick pattern of joined squares: 1 square = 4 sticks, 2 squares = 7 sticks, 3 squares = 10 sticks. How many sticks for 10 squares? For \(w\) squares?
Pattern: first square uses 4 sticks; every additional square uses 3 more sticks (shares one side).
For 10 squares: \(4+3\times9 = 4+27 = 31\)
A traffic signal cycles: Red(1), Yellow(2), Green(3), Yellow(4), Red(5),… Find the colour at positions 90, 190, 343, and write expressions describing each colour’s positions.
The cycle repeats every 4 positions: Red at position \(4n-3\); Yellow at positions \(4n-2\) and \(4n\); Green at position \(4n-1\) (for \(n=1,2,3,…\)).
| Position | ÷4 | Colour |
|---|---|---|
| 90 | 22 rem 2 | Yellow |
| 190 | 47 rem 2 | Yellow |
| 343 | 85 rem 3 | Green |
An “X”-shaped pattern of squares grows at each step. How many squares in Step 4, Step 10, Step 50? Write a general formula. How does the formula change for counting all vertices?
Step 1 has 5 squares; each new step adds 4 more squares (one arm in each diagonal direction).
General formula for Step \(n\): \(5+(n-1)\times4 = 4n+1\)
| Step 4 | \(4(4)+1=\mathbf{17}\) |
| Step 10 | \(4(10)+1=\mathbf{41}\) |
| Step 50 | \(4(50)+1=\mathbf{201}\) |
Numbers are written in an endless 4-column grid (1,2,3,4 / 5,6,7,8 / 9,10,11,12…). (a) Expression for each column. (b) Row & column of 124, 147, 201. (c) Number in row \(r\), column \(c\). (d) Pattern in positions of multiples of 3.
Endless 4-column number grid
(a) Let \(r\) = row number. Number in row \(r\) of:
| Column 1 | \(4(r-1)+1\) |
| Column 2 | \(4(r-1)+2\) |
| Column 3 | \(4(r-1)+3\) |
| Column 4 | \(4(r-1)+4\) |
(b) Using \(4(r-1)+c\):
| Number | Row | Column |
|---|---|---|
| 124 | 31 | 4 |
| 147 | 37 | 3 |
| 201 | 51 | 1 |
(c) Number in row \(r\), column \(c\):
(d) Multiples of 3 cycle through the columns in the repeating order 3, 2, 1, 4, 3, 2, 1, 4, … — e.g. 3(col 3), 6(col 2), 9(col 1), 12(col 4), 15(col 3), …
Other patterns: all of Column 4 is a multiple of 4; even numbers always fall in Columns 2 and 4; odd numbers always fall in Columns 1 and 3.