Question 262014: 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.
Answer by dabanfield(803)   (Show Source):
You can put this solution on YOUR website! 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.
Since the number is divisible by 10 the ones digit is 0. Let x be the tens digit and y be the hundreds digit. The number then is 100*y + 10*x + 0*1 Then we have:
1.) y = x - 1 and
2.) 100*y + 10*x + 0*1 = 52*(x+y+0)
Simplifying 2.) we have:
3.) 100*y + 10*x = 52*(x+y)
Substituting x - 1 for y from equation 1.) in equation 3.) we have:
100*(x-1) + 10*x = 52*(x+(x-1))
Solve the above for x and then calculate y = x-1
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