Chapter 5: Number Play (Numbers)

Using positive consecutive numbers (at least two of them):

• Odd numbers (3 and above) always work: $2k+1=k+(k+1)$.
• A number can be written this way exactly when it has an odd factor greater than 1. So the powers of 2 (1, 2, 4, 8, 16, 32, …) cannot be written as such sums; every other number can (e.g. $6=1+2+3$, $10=1+2+3+4$, $12=3+4+5$).
• The number of ways equals the number of odd factors greater than 1. So 15 (odd factors 3, 5, 15) has 3 ways: $7+8=4+5+6=1+2+3+4+5$; 9 has 2 ways ($4+5=2+3+4$).
• Not all even numbers: 8 cannot be written so, but 6, 10, 12, 14 can.
• With negative numbers allowed, every number works, and $0=-1+0+1$ (or $-3+-2+\cdots +3$).

Every number except the powers of 2 can be written as a sum of positive consecutive numbers (the number of ways = the number of odd factors > 1); with negatives allowed, 0 = −1 + 0 + 1 and every number works.

For 3, 4, 5, 6: $+++$: 18; $++-$: 6; $+-+$: 8; $+--$: −4; $-++$: 10; $-+-$: −2; $--+$: 0; $---$: −12.

For 5, 6, 7, 8: 26, 10, 12, −4, 14, −2, 0, −16.

All results are even, and −4, −2 and 0 appear every time. With the numbers $n,n+1,n+2,n+3$:

Every value is a multiple of 2 (even), and three of them do not depend on n at all.

All 8 results are even, and −4, −2, 0 always appear (the general values are 4n + 6, 2n, 2n + 2, −4, 2n + 4, −2, 0, −2n − 6).

All 8 values have the same parity (all even or all odd). Switching $-c$ to $+c$ changes the value by $+2c$, an even number, so the parity does not change; any expression can be reached from any other by such switches.

It is not limited to 4 numbers: for any list, all the expressions $a\pm b\pm c\pm \cdots$ have the same parity, for the same reason.

Token model: changing $+b$ to $-b$ means replacing b green tokens by b red tokens. The value changes by $2b$ (b tokens of one kind removed and b of the other kind added), and an even change never alters whether the result is odd or even.

Same parity for all 8: switching the sign of a term changes the value by twice that term (an even number). This holds for any number of terms.

• Even: 43 + 37 (odd + odd), 672 − 348 (even − even), 4 × 347 × 3 (has the factor 4), 543 − 479 (odd − odd).
• Odd: 708 − 477 (even − odd), 809 + 214 (odd + even), 119 × 303 (odd × odd), 513³ (odd × odd × odd).

Even: 43 + 37, 672 − 348, 4 × 347 × 3, 543 − 479. Odd: the other four.

Always-even expressions: $2n+6$, $10a-4b$, $n(n+1)$, $x^2+x$, $4p^2+2$.

Always even: 2a + 2b, 4m + 2n, 2u − 4v, 13k − 5k, 4k × 3j. Not always: 3g + 5h, 6m − 3n, x² + 2, b² + 1.

Even numbers are either $4p$ (multiples of 4) or $4q+2$.

• $4p+4q=4(p+q)$: a multiple of 4.
• $(4p+2)+(4q+2)=4(p+q+1)$: a multiple of 4.
• $4p+(4q+2)=4(p+q)+2$: never a multiple of 4; it leaves remainder 2 (e.g. 12 + 6 = 18, 8 + 10 = 18, 4 + 2 = 6).

So the sum is a multiple of 4 when both numbers are of the same type, just as an even + odd sum is odd.

The sum is a multiple of 4 when both are multiples of 4 or both leave remainder 2; a multiple of 4 plus a 4q + 2 number always leaves remainder 2.

Subtraction: $8a-8b=8(a-b)$, so always true (e.g. 56 − 16 = 40).

(6) Always true: LCM(9, 4) = 36 (9 and 4 have no common factor), so the number must be a multiple of 36.

(7) Sometimes true: LCM(6, 4) = 12, not 24. 24 and 48 are divisible by 24, but 12 and 36 are divisible by 6 and 4 and not by 24.

Subtraction: always; (6) always (LCM 36); (7) sometimes (LCM is only 12, e.g. 36).

Examples: 3, 8, 13, 18, 23, 28, …

Correct: (iv) 5k + 3 (k = 0, 1, 2, …) and (v) 5k − 2 (k = 1, 2, 3, …). (vi) 5k − 3 gives 2, 7, 12 (remainder 2), and (i)–(iii) are not of the right form.

Other expressions: $5k+8$ (k ≥ −1), $5k-7$ (k ≥ 2): any "multiple of 5 plus a number with remainder 3".

(iv) 5k + 3 and (v) 5k − 2; also e.g. 5k + 8 or 5k − 7.

$n+(n+1)+(n+2)+(n+3)=4n+6=34$, so $n=7$: 7, 8, 9, 10.

7, 8, 9, 10

$p-1$, $p-2$, $p-3$, $p-4$

p − 4, p − 3, p − 2, p − 1

(i) Sometimes: 2 + 4 = 6 is; 2 + 6 = 8 is not.

(ii) Sometimes: 20 is divisible by neither; but 27 is not divisible by 18 yet is divisible by 9.

(iii) Sometimes: 7 + 11 = 18 is divisible by 6; 7 + 8 = 15 is not.

(iv) Always: $6a+9b=3(2a+3b)$.

(v) Sometimes: 18 + 9 = 27 is; 6 + 3 = 9 is; 12 + 3 = 15 is not. ($6a+3b=3(2a+b)$ is a multiple of 9 only when $2a+b$ is a multiple of 3.)

(i) sometimes (ii) sometimes (iii) sometimes (iv) always (v) sometimes

The number minus 2 is a multiple of both 3 and 4, i.e. of 12: 2, 14, 26, 38, 50, …, all given by 12n + 2 (n = 0, 1, 2, …).

2, 14, 26, 38, …: 12n + 2

The number minus 1 is a multiple of 2, 3 and 5, i.e. of 30: 31, 61, 91, … Of these (below 100), only 91 is a multiple of 7 ($91=7\times 13$).

91 pebbles

$(6a+2)+(6b+2)+(6c+2)=6(a+b+c)+6=6(a+b+c+1)$. Yes, always (e.g. 2 + 8 + 14 = 24).

Yes: the sum is 6(a + b + c + 1).

Write $4779=7p+5$ and $661=7q+3$.

(i) $7(p+q)+8=7(p+q+1)+1$: remainder 1.

(ii) $7(p-q)+2$: remainder 2.

Visually: the full rows of 7 dots add or cancel, and only the leftover 5 and 3 dots matter: 5 + 3 = 8 = one more row and 1 dot; 5 − 3 = 2 dots.

(i) 1 (ii) 2

Each remainder is one less than the divisor, so the number plus 1 is a multiple of 3, 4 and 5, i.e. of LCM = 60. Smallest: $60-1=$ 59. Any smaller number plus 1 would be a common multiple of 3, 4, 5 smaller than 60, which does not exist.

59 (= 60 − 1, since n + 1 must be a common multiple of 3, 4, 5)

Write the number as $\cdots +1000d+100c+10b+a$.

• 2 and 5: $10b,100c,1000d,\dots$ are multiples of 10 (hence of 2 and 5), so divisibility depends only on the units digit a: even for 2; 0 or 5 for 5.
• 4: $100c,1000d,\dots$ are multiples of 100, hence of 4. So only $10b+a$, the number formed by the last two digits, matters.
• 8: $1000d,10000e,\dots$ are multiples of 1000, hence of 8. So only the last three digits $100c+10b+a$ matter.

All higher place values are multiples of 10 (for 2, 5), 100 (for 4) or 1000 (for 8), so only the last 1, 2 or 3 digits decide.

All four are correct. A number and its digit sum always leave the same remainder when divided by 9 (the number = digit sum + a multiple of 9). So one is divisible by 9 exactly when the other is.

All four are true: a number and its digit sum have the same remainder on division by 9.

Digit sums: 6, 9, 32, 28, 30. Only (ii) 405 has a digit sum divisible by 9.

Only 405.

The digit sum must be a multiple of 9; even digits (0, 2, 4, 6, 8) always give an even sum, so the digit sum must be at least 18. Two even digits give at most 16, so we need three digits. The smallest three-digit number with even digits adding to 18 is 288.

288

$6000=9\times 666+6$, so the neighbours are 5994 and 6003. 6003 is closer (3 away).

6003

From $4302=9\times 478$ to $4392=9\times 488$: $488-478+1=$ 11.

11

Every power of 10 is 1 more than a multiple of 3 ($10=9+1$, $100=99+1$, …). So a number equals its digit sum plus a multiple of 3, and leaves the same remainder as its digit sum when divided by 3.

Each power of 10 leaves remainder 1 on division by 3, so the number and its digit sum have the same remainder.

$462$: excess (units and hundreds) $=2+4=6$, short (tens) $=6$; difference $=0$, so 462 is divisible by 11 ($462=11\times 42$).

Method: (sum of digits in the 1st, 3rd, 5th, … places from the right) − (sum of digits in the 2nd, 4th, … places). If this is 0 or a multiple of 11, the number is divisible by 11; otherwise it gives the remainder (adding 11 if it is negative).

If the difference is 11 or a multiple of 11, the remainder is 0.

Yes, 462 = 11 × 42; alternate digit sums differ by 0 or a multiple of 11 exactly when the number is divisible by 11.

(i) $8-5+1=4$: remainder 4 (ii) $1-4+8=5$: remainder 5 (iii) $1-8+4=-3$: remainder 8 (iv) $9-2+5-5=7$: remainder 7 (v) $4-0+9-0+9=22$: divisible (vi) $6-7+0-7+5-8=-11$: divisible

The alternating-sign method is the same: the '+' digits are exactly the "excess" places and the '−' digits the "short" places.

(i) 4 (ii) 5 (iii) 8 (iv) 7 (v) divisible (vi) divisible; the two methods are the same.

Quick way: use the shortcuts (last digit for 2, 5, 10; last two digits for 4; last three for 8; digit sum for 3 and 9; alternating sum for 11), and get 6 from "2 and 3".

See the table (e.g. 990 is divisible by all except 4 and 8).

38: even but digit sum 11 → not divisible by 6. 225: odd → not. 186: even, digit sum 15 → divisible (186 = 6 × 31). 64: digit sum 10 → not.

$24=2^3\times 3$. Since 3 and 8 have no common factor, a number divisible by both contains $2^3$ and 3, i.e. 24. But 4 = 2² and 6 = 2 × 3 together only guarantee $2^2\times 3=12$ (their LCM), e.g. 12 and 36.

Only 186; 3 and 8 are co-prime with product 24, while 4 and 6 only force their LCM 12.

The digital root is the remainder on division by 9 (with 9 instead of 0 for multiples of 9).

(i) 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698

(ii) 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691, 700

(iii) 606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696

Twelve consecutive numbers (e.g. 20–31): 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4: the roots cycle through 1 to 9.

• Multiples of 3: 3, 6, 9, 3, 6, 9, …
• Multiples of 4: 4, 8, 3, 7, 2, 6, 1, 5, 9, then repeat.
• Multiples of 6: 6, 3, 9, 6, 3, 9, …
• One more than a multiple of 6 (7, 13, 19, 25, …): 7, 4, 1, 7, 4, 1, …

Each pattern repeats because adding a fixed number k adds k to the digital root (going round 1 to 9, like a clock).

The digital root is the remainder on division by 9 (9 for 0). Lists as above; multiples of 3: 3, 6, 9; of 4: 4, 8, 3, 7, 2, 6, 1, 5, 9; of 6: 6, 3, 9; 6k + 1: 7, 4, 1.

Every digit is 1 (the smallest odd digit), and the number of digits, their sum and the digital root are all 9 (the largest odd single digit). So the number is 11,11,11,111 (eleven crore eleven lakh eleven thousand one hundred eleven).

11,11,11,111

Adding 10 adds 1 to the remainder on division by 9: the digital root becomes 6 (e.g. 40000001 → 40000011).

6

Adding 11 adds 2 to the digital root each time (11 leaves remainder 2 on division by 9). From 10: 1, 3, 5, 7, 9, 2, 4, 6, 8, 1, 3, … All nine roots appear, in steps of 2, and the cycle repeats every 9 terms.

They go up by 2 each time (mod 9): e.g. 1, 3, 5, 7, 9, 2, 4, 6, 8, then repeat.

$9a+36b$ is a multiple of 9 and $13$ has digital root 4, so the digital root is 4.

4

(i) No relation: 13 (odd) has root 4 (even), while 22 (even) has root 4 too; 12 (even) has root 3 (odd).

(ii) Remainder on division by 9 = digital root (0 if the root is 9). Remainder on division by 3: root 1, 4, 7 → 1; root 2, 5, 8 → 2; root 3, 6, 9 → 0.

(i) No relation (ii) remainder ÷ 9 = root (0 for 9); ÷ 3: roots 1, 4, 7 → 1; 2, 5, 8 → 2; 3, 6, 9 → 0.

(i) $1+B$ ends in 0, so B = 9 (carry 1); then $A+1+1=9$, A = 7: $71+19=90$.

(ii) The tens give $A+3+\text{carry}=6$. A carry is needed, so A = 2 and $B+7=12$, B = 5: $25+37=62$.

(iii) $3\times ON=PO$. A computer check gives three answers: $31\times 3=93$, $24\times 3=72$ and $17\times 3=51$. (So O = 3, N = 1, P = 9; or O = 2, N = 4, P = 7; or O = 1, N = 7, P = 5.)

(iv) $3\times QR=PRR$: $3R$ ends in R only for R = 0 or 5. If R = 0, then $3\times Q0=P00$ would need $3Q$ to end in 0, which is impossible for a digit Q from 1 to 9. So R = 5: $3\times Q5=P55$ → Q = 8, R = 5, P = 2: $85\times 3=255$.

(i) 71 + 19 = 90 (ii) 25 + 37 = 62 (iii) 31 × 3 = 93 (also 24 × 3 = 72, 17 × 3 = 51) (iv) 85 × 3 = 255

(v) No: $13\times 8=104$ has 3 digits. The only answer is $12\times 8=96$.

(vi) The multiplier must equal the units digit of GH, and K must differ from G and H. Only 24 × 4 = 96 works (11 × 9 repeats a digit; 16 × 6 = 96 makes K = H; the others have the wrong multiplier).

(vii) B = 1, and Y is even and at most 6; Y × 6 must give the tens digit A while E × 6 ends in Y. Checking, the only solution is 105 × 6 = 630 (B = 1, Y = 0, E = 5, R = 6, A = 3).

(v) No (13 × 8 has 3 digits); 12 × 8 = 96 (vi) 24 × 4 = 96 (vii) 105 × 6 = 630

(i) $3T$ ends in T → T = 0 or 5; the answer is $50\times 3=150$ (U = 5, T = 0, P = 1).

(ii) $19\times 5=95$ (A = 1, B = 9, C = 5).

(iii) Two answers: $124\times 2=248$ (L = 1, N = 4, P = 8) and $125\times 2=250$ (L = 1, N = 5, P = 0).

(iv) $23\times 4=92$ (X = 2, Y = 3, Z = 9).

(v) QQ must be 11 (anything bigger makes a 4-digit product): $22\times 11=242$, $33\times 11=363$, $44\times 11=484$ (P = 2, 3 or 4 and R = 2P).

(vi) KKK = 111 × K is a multiple of 6 only for K = 4 (444) among the possibilities: $74\times 6=444$ (J = 7, K = 4).

(i) 50 × 3 = 150 (ii) 19 × 5 = 95 (iii) 124 × 2 = 248 or 125 × 2 = 250 (iv) 23 × 4 = 92 (v) 22 × 11 = 242 (also 363, 484) (vi) 74 × 6 = 444

Digit sum $=9+z$ must be a multiple of 9, so $z=0$ or $z=9$ (3105 and 3195). There are two answers because z can be 0 or 9: both add a multiple of 9 to the digit sum.

z = 0 or 9

$(12n+8)+(12m-4)=12(n+m)+4$. This is 16 when $n+m=1$ (a multiple of 8), but 28 when $n+m=2$ (not). So the claim is false (only sometimes true): e.g. $20+8=28$.

False: the sum is 12k + 4, which is a multiple of 8 only when k is odd (e.g. 20 + 8 = 28 is not).

$3m+3n=3(m+n)$ is a multiple of 6 exactly when $m+n$ is even, i.e. when both multiples of 3 are even (both multiples of 6, e.g. 6 + 12) or both are odd (e.g. 3 + 9). If one is even and the other odd (e.g. 6 + 9 = 15), it is not.

When both are even or both are odd multiples of 3.

(i) True: reversing does not change the digit sum (e.g. 189 → 981).

(ii) Yes, any shuffle of the digits keeps the digit sum, so it is still a multiple of 9 (189 → 819, 918, 198, …).

(i) Yes (ii) Yes, every rearrangement keeps the digit sum.

It must be even (b even) and have a digit sum $17+a+b$ divisible by 9, so $a+b=1$ or $10$.

Pairs: (a, b) = (1, 0), (8, 2), (6, 4), (4, 6), (2, 8).

(1, 0), (8, 2), (6, 4), (4, 6), (2, 8)

Divisible by 4: the last two digits q8 must be a multiple of 4, so q is even. Divisible by 11: $(8+7+3)-(q+p)=18-(p+q)$ must be 0 or ±11, so $p+q=7$ (or 18 or 29, but p + q = 18 would need q = 9 with p = 9, and q must be even; 29 is too large).

Pairs: (p, q) = (7, 0), (5, 2), (3, 4), (1, 6).

(7, 0), (5, 2), (3, 4), (1, 6)

2, 3, 4 works. If n works, so does n + 12 (LCM of 2, 3, 4 is 12): 14, 15, 16; 26, 27, 28; … They occur once every 12 numbers (first numbers 2, 14, 26, 38, …).

2, 3, 4; then every 12: 14, 15, 16; 26, 27, 28; …

A multiple of 36 is divisible by 4 and 9. $45000=36\times 1250$ exactly, so add 36 each time: 45036, 45072, 45108, 45144, 45180.

E.g. 45036, 45072, 45108, 45144, 45180

$5p-4$, $5p-2$, $5p+2$, $5p+4$

5p − 4, 5p − 2, 5p + 2, 5p + 4

Divisible by 15: last digit 0 or 5 (not 0, since the reverse would start with 0) and digit sum a multiple of 3. Reverse divisible by 6: the first digit must be even. Example: 200025 (digit sum 9; reversed 520002 is even with digit sum 9).

E.g. 200025 (others: 400035, 600045, …)

False: if $n=11k$, then $2n=11\times 2k$, always a multiple of 11.

False: 2 × 11k = 11 × 2k always.

(i) Always: $6a\times 3b=18ab=9\times 2ab$.

(ii) Always: $(2n-2)+2n+(2n+2)=6n$.

(iii) Always: swapping digits keeps the digit sum (still divisible by 3) and the last digit f stays the same (still even).

(iv) Never: it simplifies to $12b-28=12(b-2)-4$, which always leaves remainder 8 on division by 12.

(i) always (ii) always (iii) always (iv) never

Look at the remainders on division by 3. The sum is divisible by 3 exactly when the remainders add up to 0, 3 or 6, i.e. when all three remainders are the same (0, 0, 0 / 1, 1, 1 / 2, 2, 2) or all three are different (0, 1, 2). Otherwise it is not.

When all three numbers leave the same remainder, or all three leave different remainders (0, 1, 2), on division by 3.

• Two: yes, one of them is even.
• Three: yes, a multiple of 6: one is even and one is a multiple of 3.
• Four: a multiple of 24: among them are two even numbers, one of which is a multiple of 4 (giving a factor 8), and one multiple of 3.
• Five: a multiple of 120: as for four, plus one is a multiple of 5.

Two: ×2; three: ×6; four: ×24; five: ×120.

(i) $37\times 3=111$ (E = 3, F = 7, G = 1)

(ii) $575\times 5=2875$ (W = 5, O = 7, M = 2, E = 8)

(i) 37 × 3 = 111 (ii) 575 × 5 = 2875

Every multiple of 32 is a multiple of 8, and every multiple of 8 is a multiple of 4. So the sets are nested: multiples of 4 outside, multiples of 8 inside it, multiples of 32 innermost: diagram (iv).

(iv): 32s inside 8s inside 4s.