Mental Math & Number Tricks

Why A? And why not the others?

Correct answer: A. Add up all the digits (4+7+3+1 = 15); since 15 is divisible by 3, the whole number 4,731 is divisible by 3 too

Adding up all of a number's digits and checking whether that sum is divisible by 3 is the standard test for divisibility by 3, and it works here because 4+7+3+1 equals 15, which is a multiple of 3, so 4,731 is too. The option that looks at the last digit and calls it even is wrong on two counts: 4,731 actually ends in an odd digit, and even/odd only ever tells you about divisibility by 2, never by 3. The option that reaches the same correct digit sum of 15 but then concludes divisibility by 9 confuses two different rules -- a digit sum divisible by 3 only guarantees divisibility by 3, not the stricter divisibility by 9, which needs the digit sum itself to be a multiple of 9. The option using an alternating sum of digits is applying the divisibility-by-11 test to the wrong number, and even that test's own result (1) would correctly say 4,731 fails it -- it just has nothing to do with divisibility by 3.

Source: National curriculum in England: mathematics programmes of study, Key Stage 2 (GOV.UK); Wolfram MathWorld: 'Divisibility Rules'

Why B? And why not the others?

Correct answer: B. No -- the digit sum is 15, and although 15 is divisible by 3, it is not divisible by 9, so 4,731 is not a multiple of 9 even though it is a multiple of 3

4,731 is not divisible by 9, even though it is divisible by 3, because the two rules require a different strength of digit sum: divisibility by 3 only needs the digit sum to be a multiple of 3, while divisibility by 9 needs that same digit sum to be a multiple of 9. Here the digit sum is 4+7+3+1=15, which is a multiple of 3 but not of 9 (9x1=9, 9x2=18, and 15 sits between them), so the stricter test fails. The option claiming that any multiple of 3 is automatically a multiple of 9 gets the logic backwards: because 9 is 3x3, every multiple of 9 is guaranteed to be a multiple of 3, but the reverse is not true, and 4,731 is exactly the kind of number that shows why. The option that accepts the same digit sum of 15 as sufficient for divisibility by 9 skips the actual requirement that the digit sum itself be a multiple of 9. The option pointing at the last digit, 1, is testing an irrelevant property; divisibility by 9 depends on the sum of all the digits, not on any single digit in isolation.

Source: National curriculum in England: mathematics programmes of study, Key Stage 3 (GOV.UK); Wolfram MathWorld: 'Divisibility Rules'

Why C? And why not the others?

Correct answer: C. Starting from the rightmost digit, alternately subtract and add each digit (4 − 9 + 5 = 0); because the result is 0 -- itself a multiple of 11 -- the original number 594 is divisible by 11

Alternately subtracting and adding the digits of 594 from the right (4 − 9 + 5 = 0) gives a result that is itself a multiple of 11 (zero counts), which is exactly the standard test for divisibility by 11, confirming that 594 divided by 11 equals 54. The option that instead sums all the digits (5+9+4=18) is applying the digit-sum test used for divisibility by 3 and 9, not 11, and following it here would wrongly suggest the number fails since 18 is not a multiple of 11. The option that doubles the last digit and subtracts it from the rest is the standard test for divisibility by 7, not 11, and applying it to 594 (59 − 8 = 51) again leads to the wrong conclusion since 51 is not a multiple of 11 either. The option relying on the number being a palindrome is a common but false pattern: while it's true that many small multiples of 11 happen to read the same forwards and backwards (11, 22, 121), 594 is not a palindrome yet is still perfectly divisible by 11, which disproves the idea that palindrome-ness has anything to do with the actual rule.

Source: Wolfram MathWorld: 'Divisibility Rules'; Brilliant.org Wiki: 'Divisibility Rules'

Why D? And why not the others?

Correct answer: D. 4,225, from multiplying the tens digit by the next number up (6 x 7 = 42) and attaching 25

For any number ending in 5, squaring it always produces a result ending in 25, with the digits in front found by multiplying the tens digit by the next whole number up. For 65, the tens digit is 6, so 6x7=42, and attaching 25 gives 4,225, which matches exactly what long multiplication confirms (65x65=4,225). The option using 6x6=36 instead makes the common mistake of squaring the tens digit by itself rather than multiplying it by the number one greater. The option using 7x7=49 makes the opposite mistake, rounding the tens digit up before multiplying instead of pairing it with the next number. The option that correctly computes 6x7=42 but then attaches 35 instead of 25 breaks the one fixed part of this shortcut: the last two digits of the square of any number ending in 5 are always 25, never anything else.

Source: Math Fun Facts, Harvey Mudd College Department of Mathematics: 'Squares Ending in 5'

Why A? And why not the others?

Correct answer: A. 423, from computing 47 x 10 = 470 and then subtracting 47 to get 423

Multiplying by 9 is the same as multiplying by 10 and then subtracting the original number once, because 9 times any number equals 10 times that number minus one more copy of it. For 47, that means 47x10=470, and then 470−47=423, which is the correct answer. The option that stops after the x10 step and never subtracts anything is only calculating 47x10, not 47x9. The option that subtracts 37 instead of 47 has transposed the digits of the number being subtracted, taking away too little. The option that adds 47 to 470 instead of subtracting it moves in the wrong direction entirely, computing something closer to 47x11 rather than 47x9.

Source: cut-the-knot.org: 'Multiplication by 9, 99, 999, etc.'

Why B? And why not the others?

Correct answer: B. 748, from adding the digits (6+8=14), placing the 4 in the middle, and carrying the 1 into the leading digit to make it 7

To multiply a two-digit number by 11, add its two digits together and place that sum between them -- but when the sum is 10 or more, the extra ten has to be carried into the leading digit rather than just written down as a two-digit chunk. For 68, the digits sum to 6+8=14; since that's 10 or more, the 4 goes in the middle and the carried 1 is added to the leading 6, making it 7, for a final answer of 748. The option that simply inserts the full two-digit sum '14' between the original digits without carrying anything produces a number with too many digits and never actually carries the extra ten where it belongs. The option that keeps only the ones digit of the sum (4) but forgets to carry the ten into the leading digit misses exactly the step that matters once the digit sum reaches double figures. The option that does carry the 1 into the leading digit but reports the middle digit as 5 instead of 4 has simply added the two original digits incorrectly.

Source: Math Fun Facts, Harvey Mudd College Department of Mathematics: 'Multiplication by 11'

Why C? And why not the others?

Correct answer: C. 294, from adding 30% (three lots of 10%, or 252) to the full 5% block (half of one 10% block, or 42)

Breaking 35% into 30% plus 5% turns a single awkward percentage into two easy building blocks: 10% of 840 is found by moving the decimal point, giving 84, so 30% is three of those (84x3=252), and 5% is exactly half of that 10% block (42); adding 252+42 gives 294. The option landing on 336 comes from mistakenly treating 30% plus 10% as if it added up to 35%, when that combination is actually 40% of 840. The option landing on 273 comes from only adding half of the needed 5% block on top of 30%, effectively computing 32.5% instead of the full 35%. The option landing on 315 corresponds to 37.5% of 840, overshooting by adding too generous an extra sliver on top of the 30% block instead of the correct 5%.

Source: National curriculum in England: mathematics programmes of study, Key Stage 3 (GOV.UK) -- ratio, proportion and rates of change

Why D? And why not the others?

Correct answer: D. The digital roots of 47 and 38 are both 2, so the expected digital root of the product is 4; reducing 1,786's digits (1+7+8+6=22, then 2+2=4) also gives 4, so the answer is almost certainly correct, though the check can't rule out every possible error

Reducing each factor to its digital root -- repeatedly summing digits until one digit remains -- turns 47 into 2 (4+7=11, then 1+1=2) and 38 into 2 (3+8=11, then 1+1=2) as well, so the expected digital root of their true product is 2x2=4. Reducing the claimed answer, 1,786, the same way gives 1+7+8+6=22, then 2+2=4, which matches, meaning the multiplication is almost certainly correct; the check can't guarantee it's flawless, though, since casting out nines can occasionally miss specific errors, such as two digits in the final answer being swapped with each other, so a match is strong evidence of correctness rather than absolute proof. The option claiming a matching digital root proves the answer is exactly correct with no possible error overstates what the check can actually guarantee. The option that stops at the two-digit total of 22 without reducing it down to the single digit 4 never finishes the digital-root process, so comparing 22 directly against 4 produces a false 'mismatch' that isn't real. The option that suggests adding the two original numbers together (47+38) instead of finding their digital roots abandons the actual casting-out-nines procedure altogether, since the check works by comparing digital roots of the numbers being multiplied against the digital root of the product, not their sum.

Source: Wolfram MathWorld: 'Casting Out Nines'; Wikipedia: 'Casting out nines' (historical origin in Fibonacci's Liber Abaci, 1202)

Why A? And why not the others?

Correct answer: A. 9,312, from cross-subtracting to get leading digits 93, then multiplying the complements 3 x 4 = 12 for the trailing digits

The complements-from-100 shortcut multiplies two numbers close to 100 by first finding how far below 100 each one sits -- 3 for 97 and 4 for 96 -- then cross-subtracting one number's complement from the other factor (97−4, or equivalently 96−3, both give 93) to form the leading digits, and finally multiplying the two complements together (3x4=12) to form the trailing digits, combining them into 9,312. The option landing on 9,300 correctly finds the leading part but drops the complements' product entirely, effectively treating it as zero instead of 12. The option landing on 9,212 makes an error in the cross-subtraction step, computing 96−4=92 instead of pairing each number with the other's complement correctly. The option landing on 9,316 keeps the correct leading digits but miscalculates the complements' product as 16 (as if multiplying 4x4) instead of the correct 3x4=12.

Source: cut-the-knot.org: 'Product of numbers close to 100'

Why D? And why not the others?

Correct answer: D. 2,491, from computing the midpoint squared minus the squared distance (50 squared − 3 squared = 2,500 − 9)

When two numbers are the same distance from a convenient middle value, their product equals the middle value squared minus that distance squared, because (m−d)(m+d) always works out to m squared minus d squared. Here 47 and 53 both sit 3 away from the midpoint 50, so the product is 50 squared minus 3 squared, or 2,500−9=2,491, which matches direct multiplication (47x53=2,491). The option landing on 2,500 uses only the midpoint squared and forgets to subtract anything for the distance at all. The option landing on 2,494 subtracts twice the distance (2x3=6) rather than the distance squared (3 squared=9), mixing up the two different roles distance plays in the formula. The option landing on 2,509 adds the squared distance instead of subtracting it, moving in the wrong direction entirely.

Source: Wikipedia: 'Difference of two squares'; cut-the-knot.org: 'Fast Arithmetic Tips'