I am a three digit number.
I am either divisible by 3 or by 5.
I am either divisible by 4 or by 6.
I am either divisible by 5 or by 7.
I am either divisible by 6 or by 8.
I am either divisible by 7 or by 9.
I am either divisible by 9 or by 11.
What am I?
(Note: "divisible" means leaving no remainder.)
(In reply to
Puzzle Answer by K Sengupta)
Let the three-digit number be denoted by N.
At the outset, let us assume that N is divisible by 9.
Then by clues (v) and (vi), N is not divisible by either 7 or 11.
Accordingly, by clue (iii), it follows that N is not divisible by 5 and therefore by clue (i) N is not divisible by 3.
This is a contradiction since any number divisible by 9 is always a multiple of 3.
Accordingly, N cannot be divisible by 9
Then:
N is divisible by 11 ......clue (vi)
and, N is divisible by 7 .....clue (v)
=> N is not divisible by 5 ....clue(iii)
=> N is divisible by 3 .....clue (i)
By clue (ii) N is either divisible by 4 or by 6
Hence, N must be divisible by gcd(4,6)=2
=> N is divisible by lcm(2,3)=6
=> N is not divisible by 4 ......clue (ii)
and, N is not divisible by 8 .....clue (iv)
Accordingly,
N is divisible by 3,6,7,11
N is not divisible by 4,5,8
Now, 3 is a factor of 6, so we can say that N is divisible by each of 6,7 and 11 so that:
Minimum value of N = lcm(6,7,11)= 6*7*11= 462
Now, we observe that the only other 3 digit number which is a multiple of N is 462*2=924.
But 924=4*231, so that 4 divides 924.
This is a contradiction.
Consequently, the required 3-digit number is 462.
Edited on February 24, 2022, 10:17 pm
Explanation for Puzzle Answer
