The number of two digit numbers divisible by the product of the digits is: (A) 5 (B) 8 (C) 14 (D) 33
Answers
Answer: 5
Step-by-step explanation: If a number i.e(10 a+b) is divisible by product of its digits i.e (a*b) then,
(10a+b)|a &(10a+b)|b , (|----> divisible by)
∴ (10a+b)/a & (10a+b)/b are positive integer values
=> 10+(b/a) & (10a/b)+1 are +ve integer
=>10 +(b/a) & 10/(b/a)+1 are +ve integer
In above line, the bold and underlined parts are constants, so by removing them we get,
b/a is an +ve integer.
∴Value possible for b/a are 1, 2, 5 where b & a are also an +ve integer.
Now,
→Take b/a=1, ∴b=a
Two digit no.s are 11, 22, 33, 44, 55, 66, 77, 88, 99
From these only 11 satisfy the condition as 1*1=1 and 11 is divisible by 1.
→Take b/a=2 ∴b=2a
Two digits no.s are 12, 24, 36, 48
Among these 48 does not satisfy the condition.
So values are 12, 24, 36
→Take b/a=5 ∴b=5a
Two digits no.s : 15 , (No other no. satisfy the condition)
Also, 15 is divisible by their product of digits (i.e 1*5=5)
∴Collectively the no.s are 11, 12, 24, 36, 15
Total numbers = 5 (Ans:-)