Question 1200841: I am a three-digit number. my digits are 3, 5, and 7. I am divisible by 3, 5, and 7. What number am I?
Found 4 solutions by ikleyn, josgarithmetic, greenestamps, math_tutor2020:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
Any three-digit number with the digits 3, 5, and 7 is divisible by 3, since the sum of the digits 3 + 5 + 7 = 15 is divisible by 3.
In order for the number with these digits be divisible by 5, the last digit must be 5.
So, we should consider only two numbers 375 and/or 735 and check/select that number of these two which is divisible by 7.
Obviously, of these two numbers, 735 is divisible by 7, while 375 is not divisible by 7.
Hence, the answer is the number 735.
Solved.
Answer by josgarithmetic(39617)  (Show Source):
You can put this solution on YOUR website! ------------------------------------------------------------------------------
three-digit number. my digits are 3, 5, and 7. I am divisible by 3, 5, and 7
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Digit in the ones place must be 5 (because the three-digit number is to be divisible by 5). Only two possible numbers to look at: 375, or 735. The number divisible by 7 is 735.
735
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Since 3, 5, and 7 are relatively prime, any number divisible by 3, 5, and 7 is divisible by 3*5*7 = 105.
Clearly the last digit must be 5; and a quick look at 375 and 735 shows that 735 is divisible by 105 while 375 is not. So
ANSWER: 735
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
Answer: 735
Reason:
35 is a multiple of 7 (since 7*5 = 35)
The same goes for 735 (because 700+35 = 735; both 700 and 35 are multiples of 7).
735 is a multiple of 5 because it ends in 5.
735 is a multiple of 3 because the digits add to 7+3+5 = 15 which is a multiple of 3.
The prime factorization is
735 = 3*5*7^2
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