By hit and trial we find that 632 is divisible by 8. So, * must replaced by 3.
To determine if a number is divisible by 99 it needs to be divisible by 9 and 11.
Divisible by 9 test. If sum of digits is divisible by 9 then the number is divisible by 9.
Divisible by 11 test. If sum of ODD positioned digits minus the sum of the EVEN positioned digits is divisible by 11 then the number is divisible by 11.
Clearly 114345 is divisible by 9 as well as 11. so,it is divisible by 99.
Given Exp.= (1 - 1/3) (1 - 1/4) (1 - 1/5)...(1 - 1/n)
= (2/3) x (3/4) x (4/5) x ... x (n-1/n)
= 2/n.
Given Exp. ( 2 - 1/3) (2 - 3/5) (2 - 5/7) ...... (2 - 997/999)
= (5/3) x (7/5) x (9/7) x .......... x (1001/999)
= 1001/3.
Let the original number be 10p + q.
So from question,
q = 2p + 1 ...(i)
and (10q + p) - (10p + q ) = (10p + q) - 1
? 9q - 9p = 10p + q -1
? 19p - 8q = 1...(ii)
Putting the value of (i) in equation (ii) we get
19p - 8(2p + 1) = 1
? 19p - 16p - 8 = 1
? 3p = 9
? p = 3
So, q = 2 x 3 + 1 = 7
? Original number = 10 x 3 + 7 = 37
Let the two numbers be 3N and 2N
According to the question.
10 + (3N +2N ) + (3N x 2N) = 162
? 6N2 + 5N - 246 = 0
? 6N2 + 41N - 36N - 246 = 0
? N(6N + 41 ) - 6(6N + 41) = 0
? (6N + 41 ) (N - 6) = 0
? N = 6 or -41/6 (But -ve value cannot be accepted )
? Smaller number = 2N = 2 x 6 = 12 .
6 + 1 + 3 + 5 + x + 2 = 17 + x must be divisible by 9.
So, x =1.
If sum of ODD positioned digits minus the sum of the EVEN positioned digits is divisible by 11 then the number is divisible by 11.
Sum of odd positioned digits - sum of even positioned digits = (9 + 2 + 5 + 6) - (7 + 1 + x) =14 - x
It must be divisible by 11. So x = 3.
On dividing 1056 by 23, we get 21 as remainder
? Required number to be added = (23 - 21 ) = 2.
For even values of n, the number (10n - 1 ) consists of even numbers of nines and hence it will be divisible by 11.
119 is divisible by 7, 187 is divisible by 11,247 is divisible by 13 and 551 is divisible by 19. So none of the given numbers is prime.
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