Test of Mathematics Solution Subjective 74 - Sum of Squares of Digits

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.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

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.

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.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

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.