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Re: a, b, c, d, and e are five consecutive integers in increasin [#permalink]
I am a bit confused about this question.
This is how I came up with the answer

First case: let a, b, c, d, e: odd (O), even (E), odd, even, odd (1,2,3,4,5)

(A) a + b + c = O + E + O = O
(B) ab + c = E + O = O
(C) ab + d = E + E = E for example, 1*2 + 4 = 6
(D) ac + d = O + E = O
(E) ac + e = O + O = E for example, 1*3 + 5 = 8

For the second case, I start with even, odd, even, odd, even for a, b, c, d, e. (2,3,4,5,6)

(A) a + b + c = E + O + E = O
(B) ab + c = O + E = O
(C) ab + d = E + O = O
(D) ac + d = E + O = O
(E) ac + e = E + E = E

So if we start with even, then the answer is E. But if I start with odd, I got both C and E...
I'm not sure with this question maybe my solution is wrong?
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Re: a, b, c, d, and e are five consecutive integers in increasin [#permalink]
1
Expert Reply
OE


Quote:
Choice (A): a + b + c: Suppose a is an even number. Then b, the integer following a, must be odd, and c, the integer following b, must be even. Hence, a + b + c = sum of two even numbers (a and c) and an odd number (b). Since the sum of any number of even numbers with an odd number is odd (For example, if a = 4, then b = 5, c = 6, and a + b + c equals 4 + 5 + 6 = 15 (odd)), a + b + c is odd. Reject.

Choice (B): ab + c: At least one of every two consecutive positive integers a and b must be even. Hence, the product ab is an even number. Now, if c is odd (which happens when a is odd), ab + c must be odd. For example, if a = 3, b = 4, and c = 5, then ab + c must equal 12 + 5 = 17, an odd number. Reject.

Choice (C): ab + d: We know that ab being the product of two consecutive numbers must be even. Hence, if d happens to be an odd number (it happens when b is odd), then the sum ab + d is also odd. For example, if a = 4, then b = 5, c = 6, and d = 7, then ab + d = 3⋅ 5 + 7 = 15 + 7 = 23, an odd number. Reject.

Choice (D): ac + d: Suppose a is odd. Then c must also be odd, being a number 2 more than a. Hence, ac is the product of two odd numbers and must therefore be odd. Now, d is the integer following c and must be even. Hence, ac + d = odd + even = odd. For example, if a = 3, then b = 3 + 1 = 4, c = 4 + 1 = 5 (odd) and d = 5 + 1 = 6 (even) and ac + d = 3 ⋅ 5 + 6 = 21, an odd number. Reject.

Choice (E): ac + e: Suppose a is an odd number. Then both c and e must also be odd. Now, ac is product of two odd numbers and therefore must be odd. Summing this with another odd number e yields an even number. For example, if a = 1, then c must equal 3, and e must equal 5 and ac + e must equal 1 ⋅ 3 + 5 = 8, an even number. Now, suppose a is an even number. Then both c and e must also be even. Hence,

ac + e =

(product of two even numbers) + (an even number) =

(even number) + (even number) =

an even number
For example, if a = 2, then c must equal 4, and e must equal 6 and the expression ac + e equals 14, an even
number. Hence, in any case, ac + e is even. Correct.

The answer is (E).
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Re: a, b, c, d, and e are five consecutive integers in increasin [#permalink]
1
Carcass wrote:
OE


Quote:
Choice (A): a + b + c: Suppose a is an even number. Then b, the integer following a, must be odd, and c, the integer following b, must be even. Hence, a + b + c = sum of two even numbers (a and c) and an odd number (b). Since the sum of any number of even numbers with an odd number is odd (For example, if a = 4, then b = 5, c = 6, and a + b + c equals 4 + 5 + 6 = 15 (odd)), a + b + c is odd. Reject.

Choice (B): ab + c: At least one of every two consecutive positive integers a and b must be even. Hence, the product ab is an even number. Now, if c is odd (which happens when a is odd), ab + c must be odd. For example, if a = 3, b = 4, and c = 5, then ab + c must equal 12 + 5 = 17, an odd number. Reject.

Choice (C): ab + d: We know that ab being the product of two consecutive numbers must be even. Hence, if d happens to be an odd number (it happens when b is odd), then the sum ab + d is also odd. For example, if a = 4, then b = 5, c = 6, and d = 7, then ab + d = 3⋅ 5 + 7 = 15 + 7 = 23, an odd number. Reject.

Choice (D): ac + d: Suppose a is odd. Then c must also be odd, being a number 2 more than a. Hence, ac is the product of two odd numbers and must therefore be odd. Now, d is the integer following c and must be even. Hence, ac + d = odd + even = odd. For example, if a = 3, then b = 3 + 1 = 4, c = 4 + 1 = 5 (odd) and d = 5 + 1 = 6 (even) and ac + d = 3 ⋅ 5 + 6 = 21, an odd number. Reject.

Choice (E): ac + e: Suppose a is an odd number. Then both c and e must also be odd. Now, ac is product of two odd numbers and therefore must be odd. Summing this with another odd number e yields an even number. For example, if a = 1, then c must equal 3, and e must equal 5 and ac + e must equal 1 ⋅ 3 + 5 = 8, an even number. Now, suppose a is an even number. Then both c and e must also be even. Hence,

ac + e =

(product of two even numbers) + (an even number) =

(even number) + (even number) =

an even number
For example, if a = 2, then c must equal 4, and e must equal 6 and the expression ac + e equals 14, an even
number. Hence, in any case, ac + e is even. Correct.

The answer is (E).



Got it now. Thank you so much for the OE
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Re: a, b, c, d, and e are five consecutive integers in increasin [#permalink]
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