Divisibility by 3 and 9 Quiz: Digit Sum Divisibility by 3 and 9

Since N is congruent to its digit sum mod 9, and 27 = 9 times 3 is divisible by 9, N is also divisible by 9. Since every multiple of 9 is also a multiple of 3, N is divisible by both 9 and 3. Option B is false because 27 is divisible by 9. Option C is false because 27 is divisible by 3. Option D is incorrect because the digit sum test gives a definitive conclusion.

The digital root of a number equals its remainder mod 9, except when the remainder is 0 the digital root is reported as 9 rather than 0. For all nonzero remainders from 1 through 8, the digital root equals the remainder exactly. Options A through D cover remainders 1, 2, 3, and 8, all of which map to themselves as digital roots. This selection excludes remainder 0 as specified by the question.

Since N is congruent to its digit sum mod 9, divisibility of N by 9 is equivalent to divisibility of the digit sum by 9. If N is divisible by 9, its digit sum must also be divisible by 9, confirming A. Option B is false since digit sum 8 gives remainder 8, not 0. Option C is false since 12 is divisible by 3. Option D is false since divisibility of N by 3 means the digit sum is also divisible by 3.

Digit sum: 5 + 8 + 3 + 7 = 23. Now find 23 mod 3: 23 = 3 times 7 + 2, so the remainder is 2. Therefore 5,837 is congruent to 2 mod 3. Option A gives remainder 0, meaning divisible by 3, but 23 is not divisible by 3. Option B gives remainder 1. Option D gives 3, which is not a valid remainder when dividing by 3.

The answer is True. The remainder of N mod 9 equals the remainder of the digit sum mod 9. Permuting the digits of N rearranges the terms of the digit sum but does not change its total. Since the digit sum is unchanged, the remainder mod 9 is also unchanged. For example, 123, 231, and 312 all have digit sum 6 and all leave remainder 6 when divided by 9.

Since each power of 10 is congruent to 1 mod 3, replace every 10 to the power k in the expansion with 1. The expression d0 times 1 + d1 times 1 + d2 times 1 + ... simplifies to d0 + d1 + d2 + ..., which is the digit sum. So N is congruent to its digit sum mod 3. Option B gives 0, which would mean every number is divisible by 3. Options C and D retain the factor 10 or 3, which are already reduced to 1 or 0 respectively.

Digit sums: 4,356 gives 4+3+5+6 = 18, divisible by 9, confirming A. 7,416 gives 7+4+1+6 = 18, divisible by 9, confirming B. 8,901 gives 8+9+0+1 = 18, divisible by 9, confirming C. Option D: 3,451 gives 3+4+5+1 = 13, which is not divisible by 9, so 3,451 is not divisible by 9.

Digit sums: 1,300 gives 1+3+0+0 = 4, remainder 4. 2,401 gives 2+4+0+1 = 7, remainder 7. 3,112 gives 3+1+1+2 = 7, remainder 7. 4,222 gives 4+2+2+2 = 10, reducing to 1+0 = 1, remainder 1. Only 1,300 has digit sum 4, giving remainder 4 when divided by 9.

The number has seven 9s: 9 times 7 = 63. Since 63 = 9 times 7, the digit sum is divisible by 9, so 9,999,999 is divisible by 9. Option A gives 54 = 9 times 6, corresponding to six 9s. Option C gives 72 = 9 times 8, corresponding to eight 9s. Option D gives 56 which is not a multiple of 9 and does not match the digit count.

The answer is True. Since 10 is congruent to 1 mod 9, we have 10 to the power n congruent to 1 mod 9, so 10 to the power n minus 1 is congruent to 0 mod 9. The number 10 to the power n minus 1 written out is a string of n nines, for example 999, 9999, and so on, all of which are clearly divisible by 9.

Writing N = d0 + 10d1 + 100d2 + ..., since 10 is congruent to 1 mod 3, each power of 10 is also congruent to 1 mod 3. Replacing every power of 10 with 1 gives N congruent to d0 + d1 + d2 + ... mod 3. So N is divisible by 3 if and only if its digit sum is. Option B is false since powers of 10 are not multiples of 3. Options C and D are false.

Since 10 = 3 times 3 + 1, we have 10 congruent to 1 mod 3, confirming A. Raising to power k preserves the congruence giving 10 to the power k congruent to 1 mod 3, confirming B. Replacing each power of 10 with 1 in the place value expansion gives N congruent to its digit sum mod 3, confirming C. Option D is false since 10 = 6 + 4, so 10 is congruent to 4 mod 6, not 1.

Digit sums: 6,372 gives 6+3+7+2 = 18, divisible by 9. 4,206 gives 4+2+0+6 = 12, divisible by 3 but not 9. 9,126 gives 9+1+2+6 = 18, divisible by 9. 5,418 gives 5+4+1+8 = 18, divisible by 9. Only 4,206 satisfies the condition of being divisible by 3 but not 9.

Since 10 is congruent to 1 mod 3, raising both sides to the power k gives 10 to the power k is congruent to 1 to the power k which equals 1 mod 3. This constant value of 1 for all powers is what allows the digit sum test to work — every positional coefficient contributes 1 times its digit to the sum mod 3.

The answer is True. By the digit sum congruence, N is congruent to its digit sum mod 9. If the digit sum is congruent to 0 mod 9, then N is also congruent to 0 mod 9, meaning N is divisible by 9. This equivalence works in both directions — the number is divisible by 9 if and only if its digit sum is.

Digit sum: 7 + 4 + 3 = 14. Reduce again: 1 + 4 = 5. Therefore 743 is congruent to 5 mod 9 and the remainder when divided by 9 is 5. Option B gives 4, option C gives 7, option D gives 2, none of which match the correct reduced digit sum.

Since N is congruent to its digit sum mod 9 and mod 3, a digit sum divisible by 9 means N is divisible by 9, confirming A. The same reasoning for mod 3 confirms B. Every multiple of 9 is also a multiple of 3 since 9 = 3 times 3, confirming C. Option D is false — for example 6 is a multiple of 3 but not of 9 since 6 divided by 9 leaves a remainder.

Check digit sums: 8,541 gives 8+5+4+1 = 18, which is divisible by 9, so 8,541 is divisible by 9. Option B: 9+0+0+1 = 10, not divisible by 9. Option C: 1+2+2+2+3 = 10, not divisible by 9. Option D: 6+1+3+3 = 13, not divisible by 9. Only option A has a digit sum divisible by 9.

4 + 8 + 1 + 2 = 15. Since 15 = 3 times 5, it is divisible by 3, so 4,812 is also divisible by 3. Option A gives 13, option B gives 14, option D gives 16, none of which match the correct digit sum of 15.

The answer is True. If a is congruent to 1 mod m, then a to the power k is congruent to 1 to the power k which equals 1 mod m for any k. With a = 10 and m = 9, since 10 = 9 + 1 we have 10 congruent to 1 mod 9, so 10 to any power is also congruent to 1 mod 9. This is the key fact that makes the digit sum test work for 9.