When two digit number is divided by the sum of its digits the quotient is 8. If the ten's digit is diminished by three times the unit's digit the remainder is 1. What is the no.?

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Solution

let u=units digit
let t=tens digit

t-3u=1
t=3u+1

(10t+u)/(t+u)=8
10t+u=8t+8u

10(3u+1)+u=8(3u+1)+8u
30u+10+u=24u+8+8u

31u-32u=-10+8
u=2

t=3u+t=7
number: 72

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When two digit number is divided by the sum of its digits the quotient is 8. If the ten's digit is diminished by three times the unit's digit the remainder is 1. What is the no.?

let u=units digit let t=tens digit t-3u=1 t=3u+1 (10t+u)/(t+u)=8 10t+u=8t+8u 10(3u+1)+u=8(3u+1)+8u 30u+10+u=24u+8+8u 31u-32u=-10+8 u=2 t=3u+t=7 number: 72

let u=units digit
let t=tens digit

t-3u=1
t=3u+1

(10t+u)/(t+u)=8
10t+u=8t+8u

10(3u+1)+u=8(3u+1)+8u
30u+10+u=24u+8+8u

31u-32u=-10+8
u=2

t=3u+t=7
number: 72