ANSWER: October Problem of the Month

The problem, once again, was this:

Here’s the OCTOBER POTM —

Given:

A) Each of the variables x, y, and z, stands for a unique natural number from 1 thru 9 inclusive.

B) The following, therefore, represents a number in the millions place: 3,x2y,1z3

C) This number: 3,x2y,1z3 is divisible by 9.

Question: What are three other 7-digit numbers containing all of these same digits: 3, x, 2, y, 1, z, and 3 that are also divisible by 9?

ANSWER:
The correct answer was provided by Kevin Pickard, 48, a computer programmer and homeschooling parent of “two very smart girls,” ages 7 and 10. Kevin lives in Markham, ON, Canada.
Here is Kevin’s answer, in his own words:
Given that 3,x2y,1z3 is divisible by 9, then all of its digits
will add up to 9 eventually by repeated sums (eg. if x=4, y=2, z=3
then 3 + 4 + 2 + 2 + 1 + 3 + 3 = 18 and 1 + 8 = 9). So any arrangement
of these digits will still add up to 9 by repeated sums. These new
arrangements will therefore also be divisible by 9. So 3 other 7-digit
numbers that are also divisible by 9 would be as follows.
2,x1y,3z3
1,x3y,3z2
3,x3y,2z1
Plugging in the example values of x,y & z from above confirms
the result.
Josh’s note: In this problem, ANY new combination of the 7 given digits will produce a number divisible by 9. So in a sense this is a “trick problem.” The way the problem was written made some people think that only certain combinations of the digits would result in a number divisible by 9. The problem shows that the rule for divisibility by 9 is very flexible. When applying this divisibility rule, you need not take into account the order of the digits whatsoever. All that matters is that the sum of the digits is divisible by 9.


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