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Ans D - 7

2467^153∗341^72

Units digit of the product is dependent on the units digits of 2467^153 and 341^72. The units digits of 1^72 would be 1 because 1's cyclicity is 1.
Units digit of 2467^153 is 7. The cyclicity of 7 is 4.
7^0 =1
7^1 =1
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7

So, the multiplication of these two numbers would be 7*1 = 7
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153=4K+1

7^4k+1 = last digit is 7
341^72 = last digit is 1
so 7*1 =7

OA:D
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Since unit digit of 341 is 1 which would always result in unit digit of 1, on multiplying with \(2467^{153}\) it would be equal to unit digit of \(7^{153}\).
\(2467^{153} * 341^{72} = 2467^{153} * 1^{72}\)
 \(= 7^{153}\)
 \(= 7^{38*4 + 1}\)
 \(= 7^{38*4} * 7^1\)
 = 1 * 7
 = 7

Answer (D).
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cyclicity of 7 ; 7,9,3,1
unit digit of
2467^153 = 7
and 341^72 = 1
so answer IMO D ; 7

What is the unit's digit in the product 2467^153∗341^72?

(A) 0
(B) 1
(C) 2
(D) 7
(E) 9
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Quote:
What is the unit's digit in the product \(2467^{153}∗341^{72}\)?

(A) 0
(B) 1
(C) 2
(D) 7
(E) 9

\(units:2467^{153}∗341^{72}=7^{153}*1^{72}=7^{153}\)

\(cycles[7]:(7,9,3,1)=4\)

\(7^{153}…\frac{153}{4}=remainder[1]=1st.of=(7,9,3,1)=7\)

Answer (D)
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7^153* 1^72;
153/4 = 1 remainder (and 0 remainder for 72/4)
7 ^1 * 1 ^4 = 7 * 1 = 7.
Ans. D
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Cyclicity of 7 is 4 and that of 1 is 1

Hence, unit digit of 7^153 is 7 and 1^72 is 1

Unit digit of the expression is 7

D is correct
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A number with 7 in unit's digit to the powers has the cycle of numbers 7,9,3,1 respectively.
So, the sycle is 4.
153÷4=38+1.
So, 2467^153 has 7 in its unit's digit.
7×1=7
Option D

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Units first number 7 >>> Therefore we need to know the cicle of 7 >>> (7,4,1,7,4,1) >>> The cicle repeats each three numbers.

153/3= 51 >> remainder = 0 therefore the units digit of the first number is 1 since the last number of the cicle is 1 (No remainder so last number of the cicle)


Units digit second number is 1, therefore the units digit of this number to any power will be 1.

1*1=1
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What is the unit's digit in the product \(2467^{153}∗341^{72}\)??

\(...7^{153}\)

\(...7^{1}=...7\)
\(...7^{2}=...9\)
\(...7^{3}=...3\)
\(...7^{4}=...1\)
..................
\(...7^{5}=...7\)
\(...7^{6}=...9\)
\(...7^{7}=...3\)
\(...7^{8}=...1\)
It is repeated in every four terms. So,
\(...7^{149}=...7\)
\(...7^{150}=...9\)
\(...7^{151}=...3\)
\(...7^{152}\)=...1
...\(7^{153}\)=...7

\(341^{72}=...1^{72}\)=...1
Units digit of this number always ends with 1.

Units digit is 7*1=7

The answer is D.





.................
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Solution



Given
• We are given an expression: 2467^153∗341^72

To find
• The units digit of 2467^153∗341^72

Approach and Working out
• The units digit of 2467^153∗341^72 is same as the units digit of 7^153∗1^72
Units digits of 7^153∗1^72 = Units digits of 7^153∗ Units digits of 1^72
• = Units digits of 7^153 * 1
Units digits of 7^153 = Units digits of 7^(4*38 +1) = Units digits of 7^1 = 7

Hence, Units digits of 7^153∗1^72 = 7 * 1 = 7

Thus, option D is the correct answer.

Correct Answer: Option D
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Solution:

The unit digit of the product depends on the unit digit of 7^153 and 1^72

Cyclicity of 7 is 4

=>153/4 leaves a remainder of 1

=> Unit digit is 7^153 = 7

Unit digit of 341^72 would be1

=> Unit digit of the product = 7*1
= 7 (option d)
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