Expert: Josh - 7/30/2003

Question
Hi! Could you help with this quesiton.
P,Q,R, and S are distinct integers.  Find S if the following is true:

6RS
+RR4
_____
PQPQ

Answer
Hi Kim,

The key to working out this question is in recognizing what the place value represents in the metric (number) system.
These units (thousands, hundreds, tens, ones) can be regarded as multipliers (*1000,*100,*10,*1, respectively)

Translating these requirements,
we must have (6*100+R*10+S)+(R*100+R*10+4)=P*1000+Q*100+P*10+Q

simplify,
600+10R+S+100R+10R+4=1000P+100Q+10P+Q
collect like terms,
604+120R+S=1010P+101Q ...[#1]

But we also know that,
S+4=10+Q ...[#2] assuming that there is a carry
(note: this choice will become obvious when you run into trouble, if you consider the other possibility, S+4=Q, you will get a contradiction in the end.)

rewriting equation [#1],
604+120R+S=1010P+101(S-6)
604+120R+S=1010P+101S-606
1210+120R=1010P+100S ...[#3]

Hypothesis 1: Assuming that P is non-zero, then, P must be 1. Since when two 3-digit numbers are added, only 1 can overflow to create a 4-digit number.
Equation [#3] becomes,
200+120R=100S

By inspection, we find that if R=5, S=8 satisfies this equation.
Check this yourself.

Therefore, from [#2], Q=S-6=8-6=2.
The problem is solved!

P=1,R=5,Q=2,S=8.

Check: 658+554=1212. Great!

Cheers,
Josh