nonameee
Can someone please explain to me:
I think it is totally irrelevant here that two numbers are prime. In order to find the largest, we have to solve both equations to find p and q.
Am I right?
No, that' not right.
p and q are different two-digit prime numbers with the same digits, but in reversed order. What is the value of the larger of p and q?Given: \(p=prime=10x+y\) and \(q=prime=10y+x\), for some non-zero digits \(x\) and \(y\) (any two digit number can be expressed as \(10x+y\) but as both \(p\) and \(q\) are
two-digit then \(x\) and \(y\) must both be non-zero digits).
(1) p + q = 110 --> \((10x+y)+(10y+x)=110\) --> \(x+y=10\). Now, if we were not told that \(p\) and \(q\) are primes than they could take many values: 91 and 19, 82 and 28, 73 and 37, 64 and 46, ... Thus the larger number could be: 91, 82, 73, or 64. But as we are told that both \(p\) and \(q\) are primes then they can only be 73 and 37, thus the larger number equals to 73. Sufficient.
(2) p – q = 36 --> \((10x+y)-(10y+x)=
110\) --> \(x-y=4\). Again, if we were not told that \(p\) and \(q\) are primes than they could take many values: 95 and 59, 84 and 48, 73 and 37, 62 and 26, 51 and 15. Thus the larger number could be: 95, 84, 73, 62, or 51. But as we are told that both \(p\) and \(q\) are primes then they can only be 73 and 37, thus the larger number equals to 73. Sufficient.
Answer: D.
Cool and simple (after you see it set up) solution for approaching this.