Number Growth

A three-digit number grows by 9 if we exchange the second and third digits and grows by 90 if we exchange the first and second digits. By how much will it grow if we exchange the first and third digits?
Source: NCTM Mathematics Teacher, January 2006

SOLUTION
Suppose we start with 134.
Exchanging the digits 3 and 4 yields the new number 143
100+30+4=100+40+3-10+1
\textrm{-}10+1=\textrm{-}9 the number grows by 9
Exchanging the digits 1 and 3 yields the new number 314
100+30+4=300+10+4-200+20
\textrm{-}200+20=\textrm{-}180 the number grow by 180 >> 90
It looks like the digits have to be consecutive in order to have a smaller growth.
So let’s try 234.
Exchanging the digits 3 and 4 yields the new number 243
200+30+4=200+40+3-10+1
\textrm{-}10+1=\textrm{-}9 the number grows by 9 as before.
Exchanging the digits 2 and 3 yields the new number 324
200+30+4=300+20+4-100+10
\textrm{-}100+10=\textrm{-}90 the number grows by 90 as expected.
Exchanging the digits 2 and 4 yields the new number 432
200+30+4=400+30+2-200+2
\textrm{-}200+2=\textrm{-}198 the number grows by 198.
The answer seems to be 198, but we need to prove it.
Suppose a,b,c represent the three digits of a number.
Step 1: abc becomes acb
(100a+10c+b)-(100a+10b+c)=9
Simplify
9c-9b=9
c-b=1\qquad\qquad (1)
Step 2: abc becomes bac
(100b+10a+c)-(100a+10b+c)=90
Simplify
90b-90a=90
b-a=1\qquad\qquad (2)
Step 3: abc becomes cba
(100c+10b+a)-(100a+10b+c)=99c-99a
=99(c-a)
Adding Eq. (1) to Eq. (2)
c-b=1
b-a=1
————–
c-a=2
The growth equals
99(c-a)=99(2)=198

Answer: 198

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , . Bookmark the permalink.

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