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If k is the smallest positive integer such that 2,940k is the square of an integer, then k must be

A. 3
B. 5
C. 6
D. 15
E. 21


Hi,

Pls clarify whether K is a unit digit of the given No. or it is in multiplication?
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ArunpriyanJ
streamingline
If k is the smallest positive integer such that 2,940k is the square of an integer, then k must be

A. 3
B. 5
C. 6
D. 15
E. 21


Hi,

Pls clarify whether K is a unit digit of the given No. or it is in multiplication?

It's a multiplication sign.
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If k is the smallest positive integer such that 2,940k is the square of an integer, then k must be

A. 3
B. 5
C. 6
D. 15
E. 21
• Perfect square rule: A perfect square always has an even number of powers of prime factors.
That rule is often easier to remember this way:
all prime factors must come in pairs. Couplets.

There are no single "copies" of a prime factor in a perfect square.
If there is only one "copy" of a prime factor, or an odd number of copies, then the integer is not a perfect square.

• \(2,940 * k\) is a perfect square

Prime factorize: \(2,940= 2 * 2 *3*5*7*7\)
This situation will not make a perfect square.
The 3s and the 5s are not in pairs. We need one more copy of each.
(We have \(3^1\) and \(5^1\), but exponents on prime factors must be even.)

\(k\) is the number that provides the extra \(3\) and the extra \(5\)
\(k = 3*5 = 15\)

ANSWER D

\(2,940 * (3*5) = 44,100\), which is \(210^2\)
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generis
streamingline
If k is the smallest positive integer such that 2,940k is the square of an integer, then k must be

A. 3
B. 5
C. 6
D. 15
E. 21
• Perfect square rule: A perfect square always has an even number of powers of prime factors.
That rule is often easier to remember this way:
all prime factors must come in pairs. Couplets.

There are no single "copies" of a prime factor in a perfect square.
If there is only one "copy" of a prime factor, or an odd number of copies, then the integer is not a perfect square.

• \(2,940 * k\) is a perfect square

Prime factorize: \(2,940= 2 * 2 *3*5*7*7\)
This situation will not make a perfect square.
The 3s and the 5s are not in pairs. We need one more copy of each.
(We have \(3^1\) and \(5^1\), but exponents on prime factors must be even.)

\(k\) is the number that provides the extra \(3\) and the extra \(5\)
\(k = 3*5 = 15\)

ANSWER D

\(2,940 * (3*5) = 44,100\), which is \(210^2\)

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