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,

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)

good one. thanx kevin..




= (24)^(2x + 1) * (33)^(x + 1) * (17)^(x + 2) * (9)^(2x)
= 24 (336^x) * 33 (33^x) * 289 (17^x) * (81^x)
= (24 x 33 x 289) (336 x 33 x 17 x 81)^x

(24 x 33 x 289) has unit digit of 8
(336 x 33 x 17 x 81)^x has unit digit f 6 irrspective of value of x.
so (24 x 33 x 289) (336 x 33 x 17 x 81)^x has unit digit of 8.

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

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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Is my logic right ( i tried to plug some numbers and get to validate my logic. But i am not sure if this logic can be generalized)

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

this can be written in terms of unit digits as :
(4)^(2x + 1) (3)^(x + 1) (7)^(x + 2) (9)^(2x)

Then
(4)^(2x + 1) : give unit digit 4
(3)^(x + 1) : gives unit digit based 3^x and 3
(7)^(x + 2) : give unit digit based 7^x+2 and 7^2 --> 7^2 has unit digit 9
(9)^(2x) : give unit digit 1

therefore the terms can be reduced to in unit digits)
4* 3* 3^x * 7^x * 9 * 1
= 2*9 * 3^x *7^x
= 8 * 3^x *7^x
= 8 * (21)^x ( can i combine the two unit digits 3^ x * 7^ x = (21)^x = 1^x)


= 8 * 1^ x = unit digit = 8

my only doubt can i generalize this logic : 3^ x * 7^ x = (21)^x = 1^x to all unit integers.

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

2x+1=odd, so 4^odd=4 as unit

2x=even, so 9^even=1 as unit

x+1 and x+2 means that exponent of 7 is one more than exponent of 3. If we look cyclicity we always get 7 in unit when multiplying

So, 4*1*7=8 as unit

D

Basically, 17^(x+2) = 17^(x+1) * 17 (we take one 17 away from the power to get x+1 instead of x+2

Thus, 33^(x+1) x 17^(x+2) = 33^(x+1) x 17^(x+1) * 17 = (33*17)^(x+1) * 17

We can do the same with (24)^(2x + 1) * (9)^(2x) = (24*9)^(2x) * 24

From (33*17)^(x+1) * 17 we take first two unit digits, first, 3 x 7 = 21, then 1 * 7 = 7
From (24*9)^(2x) * 24 we take first two unit digits, first, 4 * 9 = 36, then 6 * 4 = 24
Finally, we have 7 and 24 or 7 * 4 = 28

Answer: 8

I hope I could help
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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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