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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.

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