<p>I need help with this math problem… x is a 2-digit positive integer. When its digits are reversed, the result is a 2-digit positive number equal to 2x + 2. What is the product of the digits of x? A. 10 B. 52 C. 54 D. 63 E. 72… This is a 5 difficulty problem. Show and explain the solution. Thanks.</p>
<p>This is a tricky problem. But think of it this way:
Let’s call the number with x’s digits reversed y. In order for y = 2x+2, the ones digit in x has to be greater than the tens digit in x. Otherwise, y would never be twice the number ‘x’. (Ex: If x = 50, the reverse of the digits would be 05).
Start thinking of numbers with bigger ones digit than tens digit. Well, if the tens digit is 1, then try 12, 13, 14, etc. None work.
Go to the tens digit as 2. 23…24…25! Yes 25 works! 2(25) + 2 = 52, and 25 switched around is 52. What’s 2 x 5? 10. A!</p>
<p>Edit: Actually, you don’t even have to try 1 in the tens digit. The reason is when the numbers are switched (12 -> 21 and 13->31 etc.), you easily see that doubling x would either be less than or greater than y.</p>
<p>I think it’s A. I just thought of possible numbers for x satisfying the 2x+2 after digit reversal.
I quickly ran through a few numbers and x turned out to be 25. When you reverse the digits you get 52 which is two times 25 plus 2.
I guess this is not an explanation really, but i hope this helps.</p>
<p>It’s 25
so the answer is 10
You know that the reversed digits gives a bigger number so that narrows it down a little because the ones digit is bigger than the tens digit
Then you know that since 2x+2 is a two digit number x has to be less than 49.
Then it’s a little bit of common sense</p>
<p>Another way to do it is narrow it down. Since the middle answer xhoice is 54, it’s obvious that either 10 or 52 is the right answer because you need to multiply a 9 and a 6 to make 54. And the number has to be lower than 49 (as I showed earlier) and theres no way to make a 2 digit number lower than 49 using 6 and 9 so it’s not c d or e. You can use the same concept to know why it’s not 52</p>
<p>Is this an actual “legitimate” CB problem or one from other prep books?</p>
<p>x is a 2-digit positive integer. When its digits are reversed, the result is a 2-digit positive number equal to 2x + 2. What is the product of the digits of x?
A. 10 B. 52 C. 54 D. 63 E. 72</p>
<p>Let x = 10m + n (m is tens’ place value, and n is units place value)
on reversing, it becomes 10n + m = 2x + 2
10n + m = 2(10m+n) + 2
10n + m = 20m + 2n + 2
8n = 19m + 2<br>
Remember that m and n are single digit numbers.</p>
<p>8n = 19m + 2
The only values satisfying the above are: (RHS must be a multiple of 8, thats how we proceed)
m=2
n = 5</p>
<p>So the product is mn = 2*5 = 10</p>
<p>^just a shortcut on solving
8n = 19m + 2:
8n - 2 = 19m
2(4n-1) = 19m
2=m and 4n-1=19.</p>
<p>============
Another way.
x=mn, where mn is not a product: m and n are digits
let y=nm
since y=2x+2=2(x+1), it’s an even number, so m is even, and n=2m or n=2m+1.
Possibilities:
24, 25
48, 49.
25 works.</p>
<p>============
The fastest way - going straight to the answer choices.
A. 10=2x5, 52=25x2+2. Done.</p>
<p>This is not an official college board question.</p>
<p>It is from:</p>
<p>Math SAT 800: How to Master the Toughest problems</p>
<p>it is very challenging</p>
<p>I agree with GCF101…</p>
<p>"The fastest way - going straight to the answer choices.
A. 10=2x5, 52=25x2+2. Done. "</p>
<p>It is crazy how often this happens: you can do a problem by a long, multi-step algebraic process. But (assuming it is a real SAT question or a good facsimile) there is ALWAYS a way to get it right without doing that process. In this case, look at some of the wrong answers:</p>
<p>Can’t be 52 because you can’t factor 52 into two 1-digit numbers.</p>
<p>You can factor 54 into 9 x 6, but then 96 is not 2 x 69 + 2…</p>
<p>But I admit, I forgot to try this method - I did the mess-around-til-I-found-the-numbers method. But at least I didn’t use algebra!</p>
<p>Math SAT 800: How to Master the Toughest problems</p>
<p>I am working through this book. It is a book of difficult problems. I’m trying to get from mid/low 600s to 750+. This book seems to be the best book for the job.</p>