If -160-we divide a two-digit number by the sum of its digits- we get - 6- as a quotient -160-and - 2- as a remainder- Now if we divide it by the product of its digits- we get - 5- as a quotient and - 2- as a remainder- Find the number- -160-

Let n(a,b) be the number n(a,b)=10a+b Given that #160; #160; n(a,b)=6(a+b)+2 10a+b=6a+6b+2 4a=5b+2 n(a,b)=5ab+2 16a+b=5ab ...

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Standard X

Mathematics

Solving a Quadratic Equation by Factorization Method

If #160;we ...

Question

If we divide a two-digit number by the sum of its digits, we get $6$ as a quotient and $2$ as a remainder. Now if we divide it by the product of its digits, we get $5$ as a quotient and $2$ as a remainder. Find the number.

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Solution

Let $n (a, b)$ be the number
$n (a, b) = 10 a + b$
Given that $n (a, b) = 6 (a + b) + 2$
$10 a + b = 6 a + 6 b + 2$
$4 a = 5 b + 2$
$n (a, b) = 5 a b + 2$
$16 a + b = 5 a b + 2$
$10 a + \frac{4 a - 2}{5} = a (4 a - 2) + 2$
$50 a + 4 a - 2 = 20 a^{2} - 10 a + 10$
$20 a^{2} - 64 a + 12 = 0$
$5 a^{2} - 16 a + 3 = 0$
$(5 a - 1) (a - 3) = 0$
$a = 3, a \neq \frac{1}{5}$
$∴$ $32$ is the required number.

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If we divide a two-digit number by the sum of its digits, we get 6 as a quotient and 2 as a remainder. Now if we divide it by the product of its digits, we get 5 as a quotient and 2 as a remainder. Find the number.

Let n(a,b) be the numbern(a,b)=10a+bGiven that n(a,b)=6(a+b)+210a+b=6a+6b+24a=5b+2n(a,b)=5ab+216a+b=5ab+210a+4a−25=a(4a−2)+250a+4a−2=20a2−10a+1020a2−64a+12=05a2−16a+3=0(5a−1)(a−3)=0a=3,a≠15∴ 32 is the required number.

If we divide a two-digit number by the sum of its digits, we get $6$ as a quotient and $2$ as a remainder. Now if we divide it by the product of its digits, we get $5$ as a quotient and $2$ as a remainder. Find the number.