A THREE DIGIT NUMBER HAS THE TENS DIGIT TWO TIMES THE HUNDREDS DIGIT. THE SUM OF THE DIGITS IS 14. IF THE THREE DIGITS ARE WRITTEN IN REVERSE ORDER THE NUMBER OBTAINED IS MORE THAN THE ORIGINAL NUMBER BY 198. FIND THE NUMBER
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Answer:
482
Step-by-step explanation:
According to question ,
Let the hundred's digit be x.
Ten's digit = 2x
Let the one's digit be y.
Original number = 100x + 10*2x + y = 120x + y
Sum of digits = 14
x + 2x + y = 14
3x + y = 14 ........ (1)
No. in reverse order = 100y + 20x + x
100y + 20x + x + 198 = 100x + 20x + y
100y - y + 198 = 100x - x
99y + 198 = 99x
y + 2 = x ............ (2)
Substituting eq. (2) in eq. (1) , we get ,
3(y + 2) + y = 14
3y + 6 + y = 14
4y = 14 - 6
4y = 8
y = 2
But we know that , x = y + 2
x = 2 + 2
x = 4
Original number = 120x + y
= 120 * 4 + 2
= 482
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