If x and y are positive integers and n = 5^x + 7^(y + 3), what is the

MANHATTAN GMAT OFFICIAL SOLUTION:

The units digit of n is determined solely by the units digit of the expressions 5^x and 7^(y + 3), because when two numbers are added together, the units digit of the sum is determined solely by the units digits of the added numbers.

Since x is a positive integer, and 5^(any positive integer) always has a units digit of 5, 5^x always ends in a 5.

However, the units digit of 7^(y + 3) is not certain, as the units digit pattern for the powers of 7 is a four-term repeat: [7, 9, 3, 1].

The question can thus be rephrased as “what is the units digit of 7^(y + 3)?”

Note: Determining y would be one way of answering the question above, but we should not rephrase to “what is y?” Because the units digits of the powers of 7 have a repeating pattern, we might get a single answer for the units digit of 7^(y + 3) despite having multiple values for y.

(1) INSUFFICIENT: This statement tells us neither the value of y nor the units digit of 7^(y + 3), as y depends on the value of x, which could be any positive integer. For example, if x = 9, then y = 2 and 7^(y + 3) has a units digit of 7. By contrast, if x = 10, then y = 4 and 7^(y + 3) has a units digit of 3.

(2) SUFFICIENT: Regardless of what multiple of 4 we pick, 7^(y + 3) will have the same units digit. Ultimately this means that n has a units digit of 5 + 3 = 8.

The correct answer is B.

Analysis :
X > 0 , Y > 0
5 raise to anything (greater than zero) will have 5 in unit digit. 7 has power cycle of 4 - 7,9,3,1
So unit digit will depend on unit digit of n will depend on 7^(y + 3), that means Unit digit of n will depend on Y.

statement 1 :
y = 2x – 16, y can be anything hence statement 1 is not sufficient.

Statement 2 :
y is divisible by 4,

Y = 4K, and Y + 3 = 4K +3

so 7 ^(Y + 3) will have unit digit of 3.

Statement 2 is sufficient.

Hence Option B is correct.

how do i know if 7^7, 7^11, 7^15 's unit are the same in the test given I only have <2 min for each q.

LAST DIGIT OF A POWER

Determining the last digit of \((xyz)^n\):

1. Last digit of \((xyz)^n\) is the same as that of \(z^n\);
2. Determine the cyclicity number \(c\) of \(z\);
3. Find the remainder \(r\) when \(n\) divided by the cyclisity;
4. When \(r>0\), then last digit of \((xyz)^n\) is the same as that of \(z^r\) and when \(r=0\), then last digit of \((xyz)^n\) is the same as that of \(z^c\), where \(c\) is the cyclisity number.

• An integer ending with 0, 1, 5, or 6, raised to any positive integer power, has the same last digit as the base.
• Integers ending with 2, 3, 7, and 8 have a cyclicity of 4.
• Integers ending with 4 have a cyclicity of 2. When the power is positive odd number, ...4^odd will end with 4, and when the power is positive even number, ...4^even will end with 6.
• Integers ending with 9 have a cyclicity of 2. When the power is positive odd number, ...9^odd will end with 9, and when the power is positive even number, ...9^even will end with 1.

Example: What is the last digit of \(127^{39}\)?

Solution: The last digit of \(127^{39}\) is the same as that of \(7^{39}\). Now, determine the cyclicity of \(7\):

1. 7^1=7 (last digit is 7)
2. 7^2=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)
5. 7^5=7 (last digit is 7 again!)
...

So, the units digit of 7^(positive integer) repeats in blocks of 4: {7, 9, 3, 1} - {7, 9, 3, 1} - ... (cyclicity of 7 in power is 4).

Now divide 39 (power) by 4 (cyclicity). The remainder is 3. So, the last digit of \(127^{39}\), which is the same as that of \(7^{39}\), is the same as the last digit of \(7^3\), which is \(3\).

Theory is here: https://gmatclub.com/forum/math-number- ... 88376.html

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Hope it helps.