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Q: What are all the two digit numbers that are divisible by both the sum and the product of its digits?

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There are five such numbers: 11, 12, 15, 24 and 36.

The digits product of two digit has number four

None. 3 digit numbers are not divisible by 19 digit numbers.

12 or 24

420

55.

12

There are 5! = 120 such numbers.

There are 5760 such numbers.

48. Assuming no digit can be used more than once, the two digit numbers divisible by 4 are: 16, 36, 48, 56, 64, 68, 84, 96 8 of them. For any number to be divisible by 4, only the last two digits need be divisible by 4; so for three digit numbers, each of the two digit numbers above can be preceded by any of the remaining 5 digits and still be divisible by 4. → 5 x 8 = 40 three digit numbers are divisible by 4 → 40 + 8 = 48 two or three digit numbers made up of the digits {1, 3, 4, 5, 6, 8, 9} are divisible by 4. If repeats are allowed, there are an extra 2 two digit numbers (44 and 88) and each of the two digit numbers can be preceded by any of the 7 digits, making a total of 7 x 10 + 10 = 80 two and three digits numbers divisible by 4 make up of digits from the given set.

from 3 digits (10x10) to 4 digits (99X99)

All whole numbers are divisible by 1. Numbers are divisible by 2 if they end in 2, 4, 6, 8 or 0. Numbers are divisible by 3 if the sum of their digits is divisible by 3. Numbers are divisible by 4 if the last two digits of the number are divisible by 4. Numbers are divisible by 5 if the last digit of the number is either 5 or 0. Numbers are divisible by 6 if they are divisible by 2 and 3. Numbers are divisible by 9 if the sum of their digits is equal to 9 or a multiple of 9. Numbers are divisible by 10 if the last digit of the number is 0.

If you mean, "What is the largest number of digits possible in the product of two 2-digit numbers" then 99 * 99 = 9801, or 4 digits. Anything down to 59 * 17 = 1003 will have 4 digits.

You had me until "product." The product of 4 digits can't be prime.

If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.

9 is the largest digit and so there are no digits that are divisible by 9.

It can have 4 digits, because the highest possible two digit numbers 99*99=9801.

There are no single-digit numbers divisible by 7000.

Every three-digit number that ends with a zero or a 5 is divisible by 5.It doesn't matter what the first 2 digits are.

575 and 757 both qualify.

3-4 :p

There are 15 of them.

None. Extremely simple proof: To be divisible by 5 the units digit must be 5. But then it would not be divisible by 2.

There are no single-digit numbers divisible by 630. 630 is divisible by 1, 2, 3, 5, 6, 7 and 9.

There are 76 such numbers. Eight more if you allow numbers to start with 0.