In two digit number the ten's place digit number is equal to the square of the unit's place digit.when we subtract 54 from the number, the number will be reversed. Find the numbers.
Answers
Answer:
The number is 93.
Step-by-step explanation:
Let the unit's digit be: x.
Ten's digit: x^2.
Original Number formed: 10x^2 + x.
Interchanged Number: 10x + x^2.
Difference between the numbers: 54.
Forming the equation, we get,
(10x^2 + x) - 54 = (10x + x^2)
{Transposing -54 and (10x + x^2) to RHS and LHS, respectively.}
→(10x^2 + x) - (10x + x^2) = 54
→10x^2 + x - 10x - x^2 = 54
→(10x^2 - x^2) + (x - 10x) = 54
→9x^2 - 9x = 54
→9(x^2 - x) = 54
{Transposing 9 to RHS.}
→x^2 - x = 54/9
→x^2 - x = 6
{By trial and error method, x ≠ 1 or 2, using x = 3, we get.}
→3^2 - 3 = 6
→9 - 3 = 6
→6 = 6.
{Hence, x = 3.}
Required Number: 10x^2 + x
→ [10 × (3^2)] + 3
→ [10 × 9] + 3
→ 90 + 3
→ 93
Checking the Solution...
Original Number: 93.
Square of unit's digit: 3^2
→9
Ten's digit: 9.
9 = 9.
{First Condition Satisfied...}
Interchanged Number: 10x + x^2
→(10 × 3) + 3^2
→30 + 9
→39
Subtracting 54 from original Number: 93 - 54
→39
39 = 39
{Second Condition Satisfied...}
Hence, required number is 93.
{Hope it helped you...}
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