I am not sure if my logic is correct, but this is what I did

I took the least gamma and delta integers. In this case, 2 and 3

2^7+3^7 and I only took the units digit in both cases

For 2^7, the units digit is 8 and for 3^7, the units digit is 7. Sum is 15. So it is Alpha

I don't think your answer will be correct. That method u use will be correct in question which ask you to find what the units digit will be. In this question logic is different.

kakz wrote:I don't think your answer will be correct. That method u use will be correct in question which ask you to find what the units digit will be. In this question logic is different.

No, looking at the units digit will work: find the units digit, and you can find the greek letter, as the five letters basically cover each "ten" twice:

If the result we're looking for ends with a 0 or a 5, it is an alpha
If the result ends with a 1 or 6, it is a beta
If the result ends with a 2 or 7, it is a gamma
If the result ends with a 3 or 8, it is a delta
If the result ends with a 4 or 9, it is a epsilon

and repeat for the next ten.

But this exact duality is what makes this problem seem unsolvable. As it turns out, it does have a single solution, but you can rest assured that nothing this complicated will come up in the GMAT. A simpler version may appear, but that too is rare.

We want to know the units digit of a gamma^7 + delta^7. We're basically looking at patterns of units digits:

If our gamma ends with a 2: 2 has a repeating pattern when raised to higher powers: 2, 4, 8, 6, and repeat. So a 2^7 will end with an 8
But if our gamma ends with a 7, we have a different pattern: 7, 9, 3, 1 and repeat. A 7^7 will end with a 3.

Likewise for a delta:

If our delta ends with a 3, there's a repeating pattern of 3,9,7,1. A 3^ ends with a 7.
But if our delta ends with an 8, then we have the pattern of 8, 4, 2, 6. An 8^7 ends with a 2.

So it seems as if the sum of these two can end with any of 4 combinations, but all of the combinations end up with the same label:

8+7 = 5 alpha
8+2=0 alpha
3+7=0 alpha
2+5-5 alpha.

so the answer is A.

kakz wrote:Starting with 0, a mathematician labels every non-negative integer as one of five types: alpha, beta, gamma, delta, or epsilon, in that repeating order as the integers increase. For instance, the integer 8 is labeled delta. What is the label on an integer that is the sum of a gamma raised to the seventh power and a delta raised to the seventh power?
(A)alpha
(B)beta
(C)gamma
(D)delta
(E)epsilon

I am getting the answer E. Plz help

hey! this one from MGMAT challange question of th week!

i did it this way --

0 1 2 3 4 5 6 7 8 9 10
a b g d e a b g d e a

now as per the question = what should we call this number

2^7+3^7 = (you certainly don't want to waste your time solving this) - Use cyclicity

2^7 -- units digit will be = 8
3^7 -- units digit will be = 7

units digit of the sum will be (8+7 = 15) i.e. 5

if you see-- all the multiple of 5 are alphas .. the answer should be A

kakz wrote:Starting with 0, a mathematician labels every non-negative integer as one of five types: alpha, beta, gamma, delta, or epsilon, in that repeating order as the integers increase. For instance, the integer 8 is labeled delta. What is the label on an integer that is the sum of a gamma raised to the seventh power and a delta raised to the seventh power?
(A)alpha
(B)beta
(C)gamma
(D)delta
(E)epsilon

I am getting the answer E. Plz help

I received a PM asking me to comment.

Just want to point out an important distinction between DS and PS questions:
If this were a DS question, we'd have to make sure that EVERY gamma� + delta� would yield the same answer.
Since this a PS problem, we have to try ONLY ONE CASE; the result must be correct for EVERY gamma� + delta�. Otherwise, the problem does not have a correct answer.

The pattern repeats in the following sequence:
1 = beta
2 = gamma
3 = delta
4 = epsilon
5 = alpha
6 = beta
etc.

One case of gamma� + delta� = 2�+3�:
2 raised to a power:
2¹ = 2
2² = 4
2³ = 8
2� = 16
2� = 32
2� = 64
2� = 128

The units digit of 3 raised to a power:
3¹ = 3
3² = 9
3³ = 27
3� = 81
3� = __3
3� = __9
3� = ___7

Adding the units digits of 2� and 3�, we get:
8+7 = 15

Thus, the units digit of 2�+3� is 5, implying that 2�+3� is a multiple of 5.
Since the sequence beta-gamma-delta-epsilon-alpha repeats in a pattern of 5, every multiple of 5 is an ALPHA.
Since 2�+3� is an alpha, ANY gamma�+delta� must also be an alpha. Otherwise, the problem does not have a correct answer.

The correct answer is A.

The Solution given by manhattan is found below. But i am having a problem in comprehending the technique of reminders used by them. I understood the technique of Geva@MasterGMAT, but would want to also understand the manhattan's technique too. So GMAT Gurus, could you please simplify the solution given by Manhattan.

Since there are five labels, given in order to all the integers, the label alpha is given to 0, 5, 10, etc. - that is, the alpha's are the multiples of 5 and end in 0 or 5. All the other labels correspond to non-multiples of 5 - in fact, they each correspond to particular remainders and particular units digits. For instance, the beta's (1, 6, 11, 16, etc.), which all end in 1 or 6, also all leave a remainder of 1 after division by 5. The gamma's correspond to a remainder of 2 (units digits = 2 or 7). Delta's correspond to a remainder of 3 (units digits = 3 or 8), and epsilon's correspond to a remainder of 4 (units digits = 4 or 9).

Now, a gamma raised to the seventh power will be large, even if we pick the smallest gamma (2 itself). But all we need is the units digit of the result. So compute the units digit in stages:

First power: units digit = 2
Second power: units digit = 2×2 = 4
Third power: units digit = 2×4 = 8 (remainder = 3)
Fourth power: units digit = 2×8 = 16 = ...6 (units digit only) (remainder = 1)
Fifth power: units digit = 2×6 = 12 = ...2 (units digit only) (remainder = 2)
Sixth power: units digit = 2×2 = 4 (remainder = 4)
Seventh power: units digit = 2×4 = 8 (remainder = 3)

Do the same for the delta.

First power: units digit = 3
Second power: units digit = 3×3 = 9 (remainder = 4)
Third power: units digit = 3×9 = 27 = ...7 (remainder = 2)
Fourth power: units digit = 3×7 = 21 = ...1 (remainder = 1)
Fifth power: units digit = 3×1 = 3 (remainder = 3)
Sixth power: units digit = 3×3 = 9 (remainder = 4)
Seventh power: units digit = 3×9 = 27 = ...7 (remainder = 2)

Gamma7 gives us a remainder of 3. Delta7 gives us a remainder of 2. Adding the remainders, we get a remainder of 5, which is the same as a remainder of 0 (remember, we're talking about division by 5).

So the sum gets a label of alpha.

The correct answer is A.

enjoylife1788 wrote:The Solution given by manhattan is found below. But i am having a problem in comprehending the technique of reminders used by them. I understood the technique of Geva@MasterGMAT, but would want to also understand the manhattan's technique too. So GMAT Gurus, could you please simplify the solution given by Manhattan.

Since there are five labels, given in order to all the integers, the label alpha is given to 0, 5, 10, etc. - that is, the alpha's are the multiples of 5 and end in 0 or 5. All the other labels correspond to non-multiples of 5 - in fact, they each correspond to particular remainders and particular units digits. For instance, the beta's (1, 6, 11, 16, etc.), which all end in 1 or 6, also all leave a remainder of 1 after division by 5. The gamma's correspond to a remainder of 2 (units digits = 2 or 7). Delta's correspond to a remainder of 3 (units digits = 3 or 8), and epsilon's correspond to a remainder of 4 (units digits = 4 or 9).

Now, a gamma raised to the seventh power will be large, even if we pick the smallest gamma (2 itself). But all we need is the units digit of the result. So compute the units digit in stages:

First power: units digit = 2
Second power: units digit = 2×2 = 4
Third power: units digit = 2×4 = 8 (remainder = 3)
Fourth power: units digit = 2×8 = 16 = ...6 (units digit only) (remainder = 1)
Fifth power: units digit = 2×6 = 12 = ...2 (units digit only) (remainder = 2)
Sixth power: units digit = 2×2 = 4 (remainder = 4)
Seventh power: units digit = 2×4 = 8 (remainder = 3)

Do the same for the delta.

First power: units digit = 3
Second power: units digit = 3×3 = 9 (remainder = 4)
Third power: units digit = 3×9 = 27 = ...7 (remainder = 2)
Fourth power: units digit = 3×7 = 21 = ...1 (remainder = 1)
Fifth power: units digit = 3×1 = 3 (remainder = 3)
Sixth power: units digit = 3×3 = 9 (remainder = 4)
Seventh power: units digit = 3×9 = 27 = ...7 (remainder = 2)

Gamma7 gives us a remainder of 3. Delta7 gives us a remainder of 2. Adding the remainders, we get a remainder of 5, which is the same as a remainder of 0 (remember, we're talking about division by 5).

So the sum gets a label of alpha.

The correct answer is A.

this is the same thing I did.