Number is 5 less than 9 times the sum of the digits. How do you find the number?

1 Answer
Oct 31, 2017

Answer:

#31#

Explanation:

Suppose that the number is #a+10b+100c+1000d+10000e+ldots# where #a,b,c,d,e,ldots# are positive integers less than #10#.

The sum of its digits is #a+b+c+d+e+ldots#

Then, according to the problem statement, #a+10b+100c+1000d+10000e+ldots+5=9(a+b+c+d+e+ldots)#

Simplify to get #b+91c+991d+9991e+ldots+5=8a#.

Recall that all variables are integers between #0# and #9#. Then, #c,d,e,ldots# must be #0#, else it is impossible for the left-hand side to add up to #8a#.

This is because the maximum value #8a# can be is #8*9=72#, while the minimum value of #91c,991d,9991e,ldots# where #c,d,e,ldots≠0# is #91,991,9991,ldots#

As most of the terms evaluate to zero, we have #b+5=8a# left.

Since the maximum possible value for #b+5# is #9+5=14#, it must be the case that #a<2#.

So only #a=1# and #b=3# work. Thus, the sole possible answer is #a+10b=31#.