Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 74 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


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Problem

The sum of squares of the digits of a three-digit positive number is 146, while the sum of the two digits in the unit’s and the ten’s place is 4 times the digit in the hundred’s place. Further, when the number is written in the reverse order, it is increased by 297. Find the number.


Solution

Let {c} is unit’s digit, {b} is ten’s digit and {a} is the hundred’s digit.

Then from the given information we get

{\displaystyle{a^2 + b^2 + c^2 = 146}} … (i)
{b + c} = {4a} …(ii)
{100a + 10b + c + 297} = {100c + 10 + a} …(iii)
{b + c} = {4a}
From (iii) we get {99(c - a)} = {297}
{\Rightarrow} {(c - a)} = {3}
Now from (ii) we get

{\Rightarrow} {b + 3} = {3a}
{\Rightarrow} {b} = {3 (a - 1)}
So b is a multiple of 3, b can be 0, 3, 6, 9.
For b = 0
not possible as
b = 0 {\Rightarrow} a = 1 {\Rightarrow} c = 4
then {\displaystyle{a^2 + b^2 + c^2 {\ne} 146}}
Similarly for b = 3 {\Rightarrow} a = 2 {\Rightarrow} c = 5
& b = 6 {\Rightarrow} a = 3 {\Rightarrow} c = 6 are not possible.
For b = 9 {\Rightarrow} a = 4 {\Rightarrow} c = 7
{\displaystyle{a^2 + b^2 + c^2 = 146}}
So the only solution is 497.