Because 9*3 = 27 and 2+7 = 9
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∙ 9y agothe answer is 9
Without digging too deeply into this question, we can see immediately that it has no solution. -- The largest possible sum of 3 digits is (9 + 9 + 9) = 27 -- 3 times (the largest possible sum of 3 digits) is (3 x 27) = 81 -- 3 times the largest possible sum of 3 digits is smaller than any 3-digit number.
No. However 54 is divisible by 3 since 54 = 3 times 18. Also, the sum of the digits of 54 is 9 and 9 = 3 times 3. (if the sum of the digits is divisible by 3, so is the number and conversely.)
The number is 45. The sum of its digits i.e. 4+5=9 Five times the sum of its digits i.e. 5 times 9 which is 45
Yes, yes, and no. 3- sum of digits must be multiple of 3. 6- sum of digits must be multiple of 3 and number must be even (multiple of 2). 9- sum of digits must be multiple of 9. (The sum of the digits here is 21.)
31
81
A number is divisible by 3 if the sum of its digits is a multiple of 3. A number is divisible by 6 if the sum of its digits is a multiple of 3 and it's even. A number is divisible by 9 if the sum of its digits is a multiple of 9.
The number is 54. The sum of its digits is 5 + 4 = 9. 54/9 = 6.
3, 6, 9
No factor of 9 has three digits. 450 is a three-digit multiple of 150 (and of 9) with a digit sum of 9.
27 2+7=9 9x3=27