<p>If a,b,c,d, and e are whole numbers and a(b(c + d) + e) is odd,then
which of the following CANNOT be even?</p>
<p>1)A
2)B
3)C
4)D
5)E</p>
<p>I choose in this problem E,but actual answer is A.
Doesn't make sense to me lets suppose c + d = even and b = odd or vice versa,then
E must be even in order to have that problem be set out to be odd,so this question
should have two solution not one!!!! It should be both A and E.
Since ODD * EVEN = EVEN,so I think this question got logical error.</p>
<p>ABB + 9B7 = AA7C</p>
<p>If A,B, and C are distinct digits in the correctly worked addition problem
above,what is the value of A + B + C ?</p>
<p>1)4
2)9
3)14
4)16
5)17</p>
<p>How is it possible to solve this without bruteforcing because I can't catch any ideas in my head to solve this without bruteforcing.</p>
<p>They have obscured the heck out of this question, but they’re testing whether you know two things: order of operations; and the fact that a product is odd only if both factors are odd.</p>
<p>If you follow order of operations, the last operation is a*[whatever the answer is to b(c+d)+e]. If the answer is odd, then a must be odd, and ALL THAT STUFF must be odd. </p>
<p>Now, ALL THAT STUFF can be odd in a couple of ways:</p>
<ol>
<li>b and (c+d) could both be odd, with e even, or</li>
<li>b or (c+d) could be even, with e odd.</li>
</ol>
<p>But the only way for a to be odd is for a to be odd. So the answer given, A, is correct.</p>
<p>(Other one later if nobody else comes along to do it.)</p>
<p>This is easier to work if you write the addition vertically, but it’s not possible (at least, I don’t think it’s possible) to make the columns line up properly in this interface.</p>
<p>The ones column suggests that either B+7=C, or B+7= something with C in the ones place. We would pronounce that number “C-teen,” and write it “1C,” but it’s value would be 10+C.</p>
<p>The tens column suggests, similarly, that B+B = 7, or B+B = 17. Neither one of those can be correct, though, since B+B = 2B, which is even. We must, then have carried 1 from the ones place. Therefore, B+7 = 10+C. And, since we carried, it’s not B+B, but rather 1+B+B that equals either 7 or 17. So B+B equals either 6 or 16, and B is either 3 or 8.</p>
<p>Look at the hundreds column. A+9 seems to have a sum of AA. If the sum is two digits, the tens digit will have to be 1. But 1+9 = 10, which is not a double-digit number like AA, with the same digit in both the tens and the ones place. Therefore, there must have been a 1 carried from the tens column, and 1+A+9 = 1+1+9 = 11, which is a double-digit number.</p>
<p>If we carried from the tens to the hundreds, then the sum of the tens column must have been 17, not 7, meaning B is 8, not 3.</p>
<p>Back to the ones column: B+7 = 8+7 = 15, so C = 5.</p>
<p>A=1; B=8; C=5. A+B+C = 1+8+5 = 14.</p>
<p>(This took me a long time. But I was typing and working at the same time. I have no clue how long it would have taken if I had been using a pencil instead of this darn keyboard. Where did you find this problem?)</p>
<p>Your solution to the first problem is impeccable. Odd * odd = even. The stuff inside the parentheses can be made odd several ways. </p>
<p>The only issue is a, which absolutely must be odd as well. </p>
<p>–</p>
<p>I don’t currently see a more direct solution to the second problem. I really like how you broke down the problem into chunks - you helped me understand the problem :)!</p>
<p>I first started out by looking at A + 9 = AA. I initially thought that A = 2, because 2 + 9 = 11, but that doesn’t completely work. Nevertheless, this provides a hint that we need to carry a one from B + B = 7 … and thus B ≠ 3, and B = 8 … and then A = 1, and then problem solved :).</p>
<p>Yes, Generic, it can be made into an even product, even if A is odd, but that is not what the question is asking. Read closely: “If a,b,c,d, and e are whole numbers and a(b(c + d) + e) is odd…” The problem is predicated on the assumption that the product a(b(c + d) + e) is odd. So it does not matter, for the purposes of this question, what happens if that product is even.</p>