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
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.
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
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.
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.
