Number Play

All sign choices have the form a±b±c±d.

Switching a term’s sign changes the value by twice that term, an even number.

For consecutive integers, a+b+c+d is even, so every sign choice keeps even parity.

Let them be n,n+1,n+2,n+3.

Their sum is 4n+6=34, so n=7.

Count backwards by one from p.

Test examples and write general multiples algebraically.

Test examples and write general multiples algebraically.

Test examples and write general multiples algebraically.

Test examples and write general multiples algebraically.

Test examples and write general multiples algebraically.

Subtract 2: the result must be divisible by both 3 and 4.

LCM(3,4)=12.

A number one more than a multiple of both 3 and 5 has form 30k+1.

It must also be odd and divisible by 7.

Candidates below 100 are 31, 61 and 91; only 91 is divisible by 7.

Write them as 6a+2, 6b+2 and 6c+2.

Their sum is 6(a+b+c)+6=6(a+b+c+1).

For the sum: 3+5=8, which leaves remainder 1 mod 7.

For 4779−661: 5−3=2.

Each remainder is one less than its divisor.

Add 1: the number becomes divisible by 3, 4 and 5.

LCM(3,4,5)=60.

If a sum has m consecutive positive integers and starts at k, it is k+(k+1)+…+(k+m−1)=mk+m(m−1)/2. For each possible length m≥2, check whether [N−m(m−1)/2] is a positive multiple of m; if so, it gives a representation.

Every odd N>1 works with m=2: N=(N−1)/2+(N+1)/2. Even numbers sometimes work, such as 12=3+4+5, but not all do: for 8, lengths 2, 3 and 4 fail the divisibility check, and length 5 or more already has a minimum sum of at least 15.

Some numbers have several representations: 15=7+8=4+5+6=1+2+3+4+5. Zero is not a sum of positive natural numbers, but if negative integers are allowed, 0=(−1)+0+1.

There are three sign positions, each with two choices, so there are 2×2×2=8 expressions.

The values are: 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.

Every result is even. Switching one sign changes the value by twice the number whose sign changed, so its parity does not change.

For the arithmetic expressions, use only whether each number is odd or even. Odd+odd and even−even are even; even×any integer is even; even−odd and odd+even are odd; odd×odd and an odd number raised to a power are odd.

Thus 43+37, 672−348, 4×347×3 and 543−479 are even. The other four are odd: 708−477, 809+214, 119×303 and 513³.

For the algebraic expressions, a factor of 2 proves an expression is always even: 2a+2b=2(a+b), 4m+2n=2(2m+n), 2u−4v=2(u−2v), 13k−5k=8k and 4k×3j=12kj.

The remaining expressions are not always even. For 3g+5h, choosing g=h=1 gives 8 (even), while g=1,h=2 gives 13 (odd). For 6m−3n, n=0 gives an even value and n=1 gives an odd value. x²+2 has the parity of x², so it is even for even x and odd for odd x. b²+1 is odd for even b and even for odd b.

An even number is either 4p (remainder 0 modulo 4) or 4p+2 (remainder 2).

Both multiples: 4p+4q=4(p+q). Neither: (4p+2)+(4q+2)=4(p+q+1). Exactly one: 4p+(4q+2)=4(p+q)+2.

(1) Always: 8a+8b=8(a+b). (2) Sometimes: 72=48+24 works, while 72=50+22 does not have both addends divisible by 8.

(3) Always: a multiple of 7 is 7j, and its multiple 7jm is divisible by 7. (4) Always: if N=12k, every factor of 12 divides N.

(5) Sometimes: 42 is divisible by 7 and 14, but not 28. (6) Always: 4 and 9 are coprime, so their product 36 divides the number.

(7) Sometimes: 12 is divisible by 4 and 6 but not 24; 24 is divisible by all three. (8) Never: odd plus even is odd, but multiples of 6 are even.

The numbers begin 3, 8, 13, 18, 23, …, increasing by 5.

They are 5k+3 for whole k≥0. They can also be written 5k−2 for k≥1, since 5k−2=5(k−1)+3.

The other listed expressions do not generate all these numbers.

Take one signed expression, for example a+b−c−d. Changing the sign before one term, such as +b to −b, changes the total by 2b, which is even.

Changing any other sign also changes the total by twice that number. Therefore, no sign change can switch an even total to an odd total or an odd total to an even total. The same reasoning works for any fixed number of terms.

So, for any fixed four integers a,b,c,d, all eight expressions a±b±c±d have the same parity. The result need not be even for arbitrary numbers; for example, 1+2+3+4 is even, so all eight are even, while 1+2+3+5 is odd, so all eight are odd.

Add each number’s digits.

A number is divisible by 9 exactly when its digit sum is divisible by 9.

The sums are 6, 9, 32, 28 and 30.

Its digit sum must be a positive multiple of 9.

A sum of even digits is even, so the least possible sum is 18.

The smallest digit multiset summing to 18 is 2,8,8; put the smallest digit first.

Digit sum of 6000 is 6, so it is 3 below the next multiple-of-9 digit sum.

6000+3=6003; the previous multiple is 5994, which is 6 below.

First is 4302=9×478; last is 4392=9×488.

Count integers 478 through 488: 488−478+1=11.

Divide each number or use the alternating-sum rule for 11.

The remainders are 4, 5, 8 and 7; the last two are exact multiples.

Each decimal place value 10ᵏ is congruent to 1 modulo 9.

So a number and its digit sum have the same remainder mod 9; repeat digit-summing to get the digital root.

10 has digital root 1.

The new number is 5+1 modulo 9, so its digital root is 6.

Adding 11 changes the number by 2 modulo 9.

Starting at digital root 1, add 2 cyclically through 1–9.

9a+36b+13=9(a+4b+1)+4.

The first term is divisible by 9; the total leaves remainder 4.

Even numbers can have different roots, for example 2 and 4.

Odd numbers can also have many different roots, for example 1, 3 and 9.

Use the last digit for 2, 5 and 10; the last two digits for 4; the last three digits for 8; the digit sum for 3 and 9; and the alternating digit sum for 11.

A number is divisible by 6 only when it is divisible by both 2 and 3.

A number is divisible by 6 exactly when it is even and its digit sum is divisible by 3.

38 is even but 3+8=11; 225 is odd; 186 is even and 1+8+6=15; 64 is even but 6+4=10.

12 is divisible by both 4 and 6, but 12 is not divisible by 24, so those two tests do not guarantee the answer.

If a number is divisible by 3 and 8, it contains the factors 3 and 2×2×2. Together these make 24.

Write a number such as 427 as 4×100+2×10+7. Since 100=99+1 and 10=9+1, the terms containing 99 and 9 are divisible by 9; only 4+2+7 affects the remainder.

Here 4+2+7=13=9+4, so the remainder is 4. The same argument works for every place value because 10ⁿ is one more than a multiple of 9.

The statements are: (i) true, (ii) true, (iii) true, and (iv) true. The first two express the same divisibility rule; the last two are their contrapositives.

10=9+1, 100=99+1, 1000=999+1, so every power of 10 leaves remainder 1 after division by 3.

When a number is expanded by place value, replace every place value by remainder 1. Its remainder modulo 3 is therefore the sum of its digits modulo 3.

A number has digital root r when it has remainder r modulo 9 (with remainder 0 written as digital root 9). Thus in 600–700 the roots repeat every 9 numbers.

For root 5 the numbers are 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698. For root 7: 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691, 700. For root 3: 603, 612, 621, 630, 639, 648, 657, 666, 675, 684, 693.

Twelve consecutive numbers show 1,2,…,9,1,2,3 in cyclic order, adjusted to the starting number. Multiples of 3 cycle through roots 3,6,9; multiples of 4 cycle through 4,8,3,7,2,6,1,5,9; multiples of 6 cycle through 6,3,9. A number one more than a multiple of 6 has a root cycling among 7,4,1.

Write a whole number as 1000k+100c+10b+a, where a is its units digit. Since 1000, 100 and 10 are multiples of 2, 5 and 10, divisibility by these numbers depends only on a.

For divisibility by 4, write the number as 100k+10b+a. The part 100k is divisible by 4, so only the last two digits, 10b+a, matter.

For divisibility by 8, write it as 1000k+100c+10b+a. The first part 1000k is divisible by 8, so only the last three digits matter.

For 158, alternate the digits from the units place: 8−5+1=4, so the remainder is 4.

For 841: 1−4+8=5, remainder 5. For 481: 1−8+4=−3, which is 8 more than a multiple of 11, so the remainder is 8. For 5529: 9−2+5−5=7, remainder 7.

For 90904 the alternating sum is 4−0+9−0+9=22, a multiple of 11. For 857076 it is 6−7+0−7+5−8=−11, also a multiple of 11.

The largest odd one-digit number is 9.

Choose 99: both digits, 9 and 9, are odd. Their sum is 18, and 1+8=9, so the digital root is 9.

Add the digits of each number: 999→27, 909→18, 900→9, 90→9 and 990→18. Each digit sum is a multiple of 9.

For any number made only of 0s and 9s, every nonzero place-value term is 9 times a whole number. Their sum is therefore a multiple of 9.

From the units place, 1 leaves remainder 1, 10 leaves remainder −1, and 100 leaves remainder 1 when divided by 11. The signs continue to alternate for successive place values.

Write 462=4×100+6×10+2. Its remainder matches 4−6+2=0, so 462 is divisible by 11.

For any number, take the alternating sum of its digits. If that result is 0 or a multiple of 11, the number is divisible by 11; otherwise it has the same remainder as that signed sum, adjusted to 0 through 10.

Use each column separately and remember that different letters stand for different digits; the first digit of a number cannot be zero.

(i) 71+19=90, so A=7, B=9. (ii) 25+37=62, so A=2, B=5.

(iii) The valid assignments are 17×3=51, 24×3=72 and 31×3=93. (iv) 85×3=255, so Q=8, R=5 and P=2.

For the multiplication examples, 12×8=96 fits PQ×8=RS; the option that fits GH×H=9K is 24×4=96; and 105×6=630 fits BYE×6=RAY.

The product has one more digit than UT and ends in T.

Checking the place-value multiplication gives 50×3=150.

The two-digit number AB times 5 must end in B.

The solution is 19×5=95.

The book’s solution gives 125×2=250.

Match corresponding digits.

The solution gives 23×4=92.

Match X=2, Y=3, Z=9.

The solution is 22×11=242.

Thus P=2, Q=1 and R=4.

The solution is 74×6=444.

The solution is 37×3=111.

The solution is 575×5=2875.

The digit sum is 3+1+z+5=9+z.

This must be a multiple of 9, and z is one digit.

Write them as 12n+8 and 12m−4.

Their sum is 12(n+m)+4, which is not always divisible by 8.

For n+m=0, the sum is 4, a counterexample (positive examples also exist).

3m+3n=3(m+n).

For this to be divisible by 6, m+n must be even.

Reordering digits leaves their sum unchanged.

The digit sum remains divisible by 9.

Divisibility by 18 requires divisibility by 2 and 9.

So b∈{0,2,4,6,8} and 17+a+b is a multiple of 9.

To be divisible by 44, the number must be divisible by both 4 and 11 because 4 and 11 have no common factor.

The last two digits are q8. Among 08, 18, 28, …, 98, the ones divisible by 4 are 08, 28, 48, 68 and 88. Thus q can only be 0, 2, 4, 6 or 8.

The alternating digit sum for 3p7q8 is 8−q+7−p+3=18−p−q. It must be a multiple of 11. Since p and q are digits and q is even, 18−p−q can be 0 or 11 (the other multiples are outside the possible range).

If q=0, p=7; if q=2, p=5; if q=4, p=3; if q=6, p=1; if q=8, p would be −1, which is not a digit.

2,3,4 works.

The conditions repeat every LCM(2,3,4)=12.

36=4×9, so test divisibility by both 4 and 9.

45036, 45072, 45108, 45144 and 45180 each pass both tests.

Consecutive even numbers differ by 2.

Move two steps of 2 either side of the middle.

Choose a number ending in 5 so it can be divisible by 5.

Use 200025: digit sum 9, so it is divisible by 3 and 5, hence by 15.

Its reverse 520002 is even and has digit sum 9, so it is divisible by 2 and 3, hence by 6.

Write any multiple as 11k.

Doubling gives 22k=11(2k).

Write the terms algebraically and factor or simplify.

Write the terms algebraically and factor or simplify.

Write the terms algebraically and factor or simplify.

Write the terms algebraically and factor or simplify.

Each remainder modulo 3 is 0, 1 or 2.

The sum is divisible by 3 exactly when the three remainders add to 0, 3 or 6.

For 2 consecutive integers, one is even, so their product is divisible by 2.

For 3 consecutive integers, one is divisible by 3 and at least one other is even. These different factors give 2×3=6.

For 4 consecutive integers, one is divisible by 4, another different integer is even, and one is divisible by 3. Thus the product contains 4×2×3=24 as factors.

For 5 consecutive integers, the factors include a multiple of 4, another even number, a multiple of 3 and a multiple of 5. Together these provide 4×2×3×5=120.

Every multiple of 32 is a multiple of 8.

Every multiple of 8 is a multiple of 4.