Hi,

\((24)^{2x + 1}*(33)^{x + 1}*(17)^{x + 2}*(9)^{2x}\)
\(=(24^2)^x*24*33^x*33*17^x*17^2*(9^2)^x\)
\(=(24^2)^x*(33*17)^x*(24*33)*17^2*(9^2)^x\)
considering only the unit digits;
\(=(6)^x*(1)^x*2*9*(1)^x\)
\(=6*1*2*9*1\)
\(=8\)

Answer (D)

Regards,

The answer is D

(24)^(2x + 1)
(2x+1) is an odd number since even + odd = odd
4 to an even power ends with a 4.

(33)^(x + 1)*(17)^(x + 2)
x could be even or odd. So if (x+1) is even, (x+2) will be odd. On the contrary, if (x+1) is odd, (x+2) will be even.
Trying some values for x:

If x =1

(33)^(2)*(17)^(3) = 9*3 = 27

If x= 2

(33)^(3)*(17)^(4) = 7*1 = 7

and so on......the table below shows the pattern:

Number
3 7
Power
2 9 9
3 7 3
4 1 1
5 3 7
6 9 9


(9)^(2x)

the power will be even, thus the units digit is 1.



Multiplying 4*7*1 = 8

In these questions, since the answer will be true for any value of x, we can choose the min. value of x (in this case 1) and solve..

(D)

Remember that in the GMAT, it pays to look for shortcuts!

33^(x+1)*17^(x+2)= ((33*17)^(x+1))*17 which has a units digit of 7!

solved in 00:34

consider x =1
24^3 * 33^2 * 17^3 * 9^2

units digit will be
4*9*3*1

=> 8

satisfies for any positive integer.

Let value of x;

x=1

24^3*33^2*17^3*9^2

unit digits = 4*9*3*1 = 108

unit digit = 8

answer = D

Hi Bunuel,

In these kind of questions where in it is asked that X is a positive integer, is substituting and value of X a good idea to solve it quickly. Though by taking X=1 or 2 i have arrived at unit's digit as 8 but will it hold for all values of X.

Thanks

Yes. There is only one correct answer in a PS question, thus every x should give the same correct answer.

Units digits, exponents, remainders problems directory: new-units-digits-exponents-remainders-problems-168569.html

Hope it helps.
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Here is another method to answer this question very quickly

Just observe the Language of the question "If x is a positive integer, what is the units digit of (24)^(2x + 1)*(33)^(x + 1)*(17)^(x + 2)*(9)^(2x)?"

The "is" part confirms that the result of this question will be unique for any value of x which is a positive Integer.

Hence this question becomes much easier for any chosen positive integer value of x,

Let's take x = 1

Now the question becomes

(24)^(2x + 1)*(33)^(x + 1)*(17)^(x + 2)*(9)^(2x) = (24)^(2 + 1)*(33)^(1 + 1)*(17)^(1 + 2)*(9)^(2)

IMPORTANT POINT : In calculation of the Unit Digit, only Unit Digit matters and all the digits other than unit digit of numbers become redundant.

But (24)^(2 + 1) will have same unit digit as 4^(2+1) i.e. 4^3 i.e. 4

and But (33)^(1 + 1) will have same unit digit as 3^(1+1) i.e. 3^2 i.e. 9

and But (17)^(1 + 2) will have same unit digit as 7^(1+2) i.e. 7^3 i.e. 3

and But (9)^(2) will be 1

i.e. Unit digit of (24)^(2x + 1)*(33)^(x + 1)*(17)^(x + 2)*(9)^(2x) = 4 x 9 x 3 x 1 = 8

Answer: Option
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Since we are only concerned with the units digit, we can simplify the expression as:

(4)^(2x + 1)*(3)^(x + 1)*(7)^(x + 2)*(9)^(2x)

This simplified expression will have the same units digit as the given expression. Next, let’s look at the units digit patterns of powers of 4, 3, 7, and 9, respectively:

Units digits of powers of 4: 4-6 (the patterns repeats in a cycle of 2 with 4^odd = 4 and 4^even = 6)

Units digits of powers of 3: 3-9-7-1 (the patterns repeats in a cycle of 4 with 3^(a multiple of 4) = 1)

Units digits of powers of 7: 7-9-3-1 (the patterns repeats in a cycle of 4 with 7^(a multiple of 4) = 1)

Units digits of powers of 9: 9-1 (the patterns repeats in a cycle of 2 with 9^odd = 9 and 9^even = 1)

Since 2x + 1 is always odd regardless of what integer x is, 4^(2x + 1) = 4^odd = 4. Similarly, since 2x is always even regardless of what integer x is, 9^(2x) = 9^even = 1. However, since x + 1 (the exponent of 3) and x + 2 (the exponent of 7) are sometimes odd and sometimes even depending on what integer x is, we are going to change tactics in analyzing the units digit of (3)^(x + 1)*(7)^(x + 2). Notice that:

(3)^(x + 1)*(7)^(x + 2) = (3)^(x + 1)*(7)^(x + 1)*7 = (3*7)^(x + 1)*(7) = (21)^(x + 1)*(7)

Since we are only concerned with the units digit, we can simplify (21)^(x + 1)*(7) as (1)^(x + 1)*(7). Since 1 raised to any power is 1, the units digit of (1)^(x + 1)*(7) or (21)^(x + 1)*(7) is 1*7 = 7. With this, we can see that the units digit of (4)^(2x + 1)*[(3)^(x + 1)*(7)^(x + 2)]*(9)^(2x) is 4*[7]*1 = 28, i.e., 8.

Answer: D
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The rightmost factor -- \(9^{2x}\) -- will have a units digit of 1 for any nonnegative value of \(x\).
Thus, the rightmost factor can be ignored.

All of the remaining exponents will be positive integers even if \(x=0\).
Thus, we can solve by letting \(x=0\) and raising the units digits of the first 3 factors to the resulting exponents:
\(4^1 * 3^1 * 7^2 = 12 * 49\) = integer with a units digit of 8.


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IMO answer is option D

Posted from my mobile device

\(24^{2x+1} * 33^{x+1} * 17^{x+2} * 9^{2x}\)

Instead of factorizing the individual numbers and taking values that have similar powers together first:

= \(24^{2x+1}*33^{x+1}*17^{x+1}*17^1*81^x\)

= \(24^{2x+1}*(33*17)^{x+1}*17^1*81^x\) [Reducing the base to minimum possible value i.e. 1]

= \(4^{2x+1}*1^{x+1}*7^1*1^x\) [Taking unit values of all the entities]

= \(4*1*7*1\) [Unit value of 4 for odd powers (2x+1) = 4]

= 28

Unit value = 8

Answer (D).
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