The Proof Hiding in Every Digit Sum Number Play

Taking any four consecutive numbers, like 3, 4, 5, 6, and inserting every possible combination of '+' and '-' signs between them gives 8 different expressions (3+4+5+6, 3+4+5-6, 3+4-5+6, and so on). Evaluating every single one always produces an EVEN result, no matter which four consecutive numbers are chosen. Switching just one sign in any expression always changes its value by an even amount (replacing +b with -b changes the total by exactly 2b), and since changing a number by an even amount never changes whether it's odd or even, every one of the 8 sign-combinations must share the exact same parity as the very first one. The same fact can also be seen through the integer token model: a positive token flipped to negative removes two positive units' worth of value, again an even change.

Adding two multiples of 4 (like 4p and 4q) always gives another multiple of 4, since 4p+4q = 4(p+q) -- provable algebraically, and equally provable by picturing p rows plus q rows of 4 objects each stacking into (p+q) rows of 4. Two even numbers that AREN'T multiples of 4 (each leaving remainder 2, like 4p+2 and 4q+2) still always sum to a multiple of 4, since their two leftover 2's combine into a complete extra group of 4: (4p+2)+(4q+2) = 4(p+q+1). This chapter's real habit, proving every pattern both algebraically and visually, turns a simple observation into something genuinely certain.

Checking claims like 'if 8 divides two numbers separately, it divides their sum' (always true, since 8a+8b=8(a+b)) versus 'if a number is divisible by 8, then 8 divides any two numbers that add up to it separately' (only sometimes true -- 72=48+24 works, but 72=50+22 doesn't) builds toward genuinely general rules: if a divides both M and N, then a divides M+N and M-N; if a divides A, then a divides every multiple of A; if a divides A, then a divides every factor of the number that A itself divides into; and if A is divisible by both k and m, then A is divisible by the LCM of k and m -- not simply by k times m, since that would overcount any factors k and m already share.

Checking whether 427 is divisible by 9 by adding its digits (4+2+7=13, then 1+3=4) works because every power of 10 is exactly 1 more than a multiple of 9: 10=9+1, 100=99+1, 1000=999+1, and so on. Splitting 7309 into 7x1000+3x100+0x10+9x1 and rewriting each power of 10 as '(a multiple of 9) + 1' shows the whole number equals (a big multiple of 9) plus exactly (7+3+0+9), its own digit sum -- meaning the number's remainder upon division by 9 is EXACTLY the same as its digit sum's remainder upon division by 9. A number is divisible by 9 exactly when its digit sum is divisible by 9, and this is not a coincidence needing memorisation, but a direct, provable consequence of how powers of 10 relate to 9.

Unlike 9's pattern, powers of 10 relative to 11 alternate: 1 is 1 more than a multiple of 11 (trivially, 0x11+1), but 10 is 1 LESS than a multiple of 11 (11-1), 100 is 1 MORE (99+1), 1000 is 1 LESS (1001-1), and so on, flipping sign at every place value. This directly gives the alternating-sum test: placing alternating '+' and '-' signs before every digit starting from the units place, then evaluating, gives exactly the number's remainder pattern relative to 11 -- for 328105, computing -3+2-8+1-0+5 = -3 confirms the number is 3 short of (or equivalently, 8 more than) a multiple of 11. Divisibility by 3 follows the exact same reasoning as divisibility by 9, just checking digit-sum divisibility by 3 instead of 9, since every power of 10 is ALSO 1 more than a multiple of 3.

Repeatedly adding a number's digits until only one digit remains gives its digital root -- for 489710, that's 4+8+9+7+1+0=29, then 2+9=11, then 1+1=2. This single digit always equals the number's remainder on division by 9 (or exactly 9 itself, if the number is a genuine multiple of 9), a direct consequence of the same divisibility-by-9 proof already established. Aryabhata II, working around 950 CE in his text Mahasiddhanta, described exactly this repeated-digit-summing method, using it specifically as a quick way to check arithmetic calculations for errors -- a technique still mathematically valid, and still occasionally useful, more than a thousand years later.

In the cryptarithm PQ x 8 = RS (a 2-digit number times 8 giving another 2-digit number), Guna reasons through it directly: 12x8=96 works, but 13x8=104 doesn't (already 3 digits), ruling out every 2-digit number above 12 immediately, without checking each one by brute force. Similarly, in BYE x 6 = RAY, Anshu reasons that B must be 1 (since B=2 would already push the product past 3 digits), and that Y must be even and less than 7 (since Y=7 would give a product exceeding 3 digits too) -- cryptarithms throughout this chapter are solved this way, through logical constraints on place value and digit range, not by testing every possible digit combination.