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Bunuel
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Among 200 integers 2^3, 2^6, 2^9, 2^12, ..., 2^600, how many numbers 12 Apr 2023, 10:45
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Among 200 integers \(2^3, 2^6, 2^9, 2^{12},…., 2^{600}\), how many numbers are there whose unit digit is 4 ?

A. \(20\)

B. \(25\)

C. \(40\)

D. \(50\)

E. \(100\)
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D
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Among 200 integers 2^3, 2^6, 2^9, 2^12, ..., 2^600, how many numbers 13 Apr 2023, 09:08
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Bunuel
Among 200 integers \(2^3, 2^6, 2^9, 2^{12},…., 2^{600}\), how many numbers are there whose unit digit is 4 ?

A. \(20\)

B. \(25\)

C. \(40\)

D. \(50\)

E. \(100\)

In the expression \(2^3, 2^6, 2^9, 2^{12},…., 2^{600}\), applying Arithmetic progression on the powers- 3,6,9...600, , one will see that there are 200 terms.

Consider the cyclicity of 2- 2,4,8,6. So out of every 4 element, 1 element will have 4 as unit digit.

Therefore out of 200 elements, 50 elements will have unit digit as 4.
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Among 200 integers 2^3, 2^6, 2^9, 2^12, ..., 2^600, how many numbers 12 Apr 2023, 12:36
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Bunuel
Among 200 integers \(2^3, 2^6, 2^9, 2^{12},…., 2^{600}\), how many numbers are there whose unit digit is 4 ?

A. \(20\)

B. \(25\)

C. \(40\)

D. \(50\)

E. \(100\)

The unit digit of the term will be 4 if the power (radicle) is in the form of 4n + 2.

Also, the powers are positive multiples of 3.

Therefore the powers which result in the unit digit = 4 must be a multiple of 3 and (4n + 2)

Hence the terms should have a general representation of 3*(4n + 2) = 12n + 6

The expression \(6 \leq 12n + 6 \leq 600\).

In a nutshell, we have to find the number of terms between 6 and 600 inclusive which are of form 12n + 6; n is a non-negative integer.

\(6 \leq 12n + 6 \leq 600\)

Subtracting 6 on all sides of the inequality

\(0 \leq 12n \leq 594\)

Dividing by 12

\(\frac{0}{12} \leq 12n \leq \frac{594}{12}\)

\(0 \leq n \leq 49.50\)

Number of terms between 0 and 49, both inclusive

(49 - 0)+1 = 50

Option D
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Among 200 integers 2^3, 2^6, 2^9, 2^12, ..., 2^600, how many numbers 13 Apr 2023, 03:38
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gmatophobia
Bunuel
Among 200 integers \(2^3, 2^6, 2^9, 2^{12},…., 2^{600}\), how many numbers are there whose unit digit is 4 ?

A. \(20\)

B. \(25\)

C. \(40\)

D. \(50\)

E. \(100\)

The unit digit of the term will be 4 if the power (radicle) is in the form of 4n + 2.

Also, the powers are positive multiples of 3.

Therefore the powers which result in the unit digit = 4 must be a multiple of 3 and (4n + 2)

Hence the terms should have a general representation of 3*(4n + 2) = 12n + 6

The expression \(6 \leq 12n + 6 \leq 600\).

In a nutshell, we have to find the number of terms between 6 and 600 inclusive which are of form 12n + 6; n is a non-negative integer.

\(6 \leq 12n + 6 \leq 600\)

Subtracting 6 on all sides of the inequality

\(0 \leq 12n \leq 594\)

Dividing by 12

\(\frac{0}{12} \leq 12n \leq \frac{594}{12}\)

\(0 \leq n \leq 49.50\)

Number of terms between 0 and 49, both inclusive

(49 - 0)+1 = 50

Option D

How do we know that the exponent has to be of the form "4n+2" in order for the unit digit to be "4"?
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Among 200 integers 2^3, 2^6, 2^9, 2^12, ..., 2^600, how many numbers 13 Apr 2023, 03:43
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Charlotte12345
How do we know that the exponent has to be of the form "4n+2" in order for the unit digit to be "4"?

The cyclicity of 2 is 4 and is shown below -

2
4
8
6

4 is at the unit's place when the power is 2 and then repeats after every interval of 4, i.e. when the power is 6, 10 ..

Therefore the general representation of such a power is 4n + 2
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Among 200 integers 2^3, 2^6, 2^9, 2^12, ..., 2^600, how many numbers 30 Oct 2024, 00:38
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