A number of three digits has a hundred digits 4 times the unit digit and the sum of three
digits are 14. If the three digits are written in the reverse order, the value of the number is
decreased by 594. Find the number.
Let’s consider the digit at tens place as x
And let the digit at unit place be y
Then, the digit at hundred place = 4y
Hence, the number is 100 x 4y + 10 × x + y × 1 = 400y + 10x + y = 401y + 10x
So, reversing the number = 100 × y + 10 × x + 4y × 1 = 100y + 10x + 4y = 104y + 10x
Then according to the first condition given in the problem, we have
x + y + 4y = 14
x + 5y = 14 … (i)
And according to the second condition given in the problem, we have
401y + 10x = 104y + 10x + 594
401y – 104y = 594
297y = 594
y = 594/297
y = 2
On substituting the value of y in equation (i), we have
x + 5(2) = 14
x + 10 = 14
x = 14 – 10
x = 4
Therefore, the number is = 10x + 401y = 10(4) + 401(2) = 40 + 802 = 842.
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