The Original

Algebra Level 1

The tens' digit of a number is 3 less than the units' digit. If the number is divided by the sum of digits, the quotient is 4 and the remainder is 3. What is the original number?


The answer is 47.

Riah Ann Fermin by Riah Ann Fermin , 31, Philippines

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2 solutions

Reineir Duran
Jan 11, 2015

3 Helpful 0 Interesting 0 Brilliant 0 Confused

t = t e n ′ s d i g i t , u = u n i t ′ s d i g i t t=ten's~digit, u=unit's~digit

t = u − 3 t=u-3 ⟹ \implies 1 \boxed{1}

10 t + u t + u = 4 + 3 t + u \dfrac{10t+u}{t+u}=4+\dfrac{3}{t+u}

Multiplying both sides by t + u t+u , we get

10 t + u = 4 t + 4 u + 3 10t+u=4t+4u+3

6 t − 3 u = 3 6t-3u=3

2 t − u = 1 2t-u=1 ⟹ \implies 2 \boxed{2}

Substituting 1 \boxed{1} in 2 \boxed{2} , we get

2 ( u − 3 ) − u = 1 2(u-3)-u=1

2 u − 6 − u = 1 2u-6-u=1

u = 7 u=7

It follows that

t = u − 3 = 7 − 3 = 4 t=u-3=7-3=4

So the original number is 47 \boxed{47} .

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