Question 27444: I'm not quite sure if this is the right section to be asking this question in, but if you can help please do.
It says to "Solve each of the following problems about three-digit numbers."
The Problem:
A three-digit number, which is divisible by 10, has a hundreds digit that is one less than its tens digit. The number also is 52 times the sum of its digits. Find the number.
Thanks for your time and help. I tried doing this problem but came up with an irrational number... the hundreds digit being -21/1860... and that can't be the answer. Thanks again
Answer by Z(8)   (Show Source):
You can put this solution on YOUR website! It may not be the most efficient way, but this is how I do problems like this:
Since it is divisible by 10, that means the last digit has to be a 0.
The hundreds digit (the first digit) is one less than the tens digit (the middle digit) This means your only possibilities would be 890, 780, 670, 560, 450, 340, 230, and 120.
Now the number has to be 52 times the sum of its digits, meaning if we add up all the digits and multiply it by 52 it will equal the number.
890; 8+9+0=17; 17 x 52=884 nope
780; 7+8+0=15; 15 x 52=780 yep, these are your numbers.
670; 6+7+0=13; 13 x 52=676 nope
560; 5+6+0=11; 11 x 52=572 nope
450; 4+5+0=9; 9 x 52=468 nope
340; 3+4+0=7; 7 x 52=364 nope
230; 2+3+0=5; 17 x 52=884 nope
120; 1+2+0=3; 17 x 52=884 nope
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