Which one of the following is not a prime number?
Answer: D
Test each for divisibility.
\[ 91 = 7 \times 13 \]
31, 61 and 71 have no divisor other than 1 and themselves, so 91 is the composite one. It is a classic trap: it is odd, does not end in 5, and its digits sum to 10, so the easy tests all pass - only 7 catches it.
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The positive integers mm and nn leave remainders of 2 and 3, respectively, when divided by 6. m>nm>n.
What is the remainder when m-nm-n is divided by 6?
Answer: C
We are given that the numbers mm and nn, when divided by 6, leave remainders of 2 and 3, respectively.
Hence, we can represent the numbers mm and nn as 6p+26p+2 and 6q+36q+3, respectively, where pp and qq are suitable integers.
Now,
m−n=(6p+2)−(6q+3)
=6p−6q−1
=6(p−q)−1
A remainder must be positive, so let’s add 6 to this expression and compensate by subtracting 6:
6(p−q)−1=6(p−q)−6+6−1
=6(p−q)−6+5
=6(p−q−1)+5
Thus, the remainder is 5
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1397 x 1397 = ?
Answer: A
\(1397\times 1397 = (1397)^{2}\)
\(= (1400-3)^{2}\)
\(= (1400)^{2}+(3)^{2}-(2\times 1400\times 3)\)
\(= 1960009 - 8400\)
\(= 1951609\)
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\(\left \{(476 + 424)^{2} - 4 \times 476 \times 424 \right \} = ?\)
Answer: C
Given Expression = \([(a + b)^{2} - 4ab]\), where \(a = 476\) and \(b = 424\)
\(=[(476 + 424)^{2} - 4 \times 476 \times 424]\)
\(= [(900)^{2} - 807296]\)
\(= 810000 - 807296\)
\(= 2704\)
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A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as the remainder. The number is:
Answer: D
\[ \text{divisor} = 555 + 445 = 1000 \]
\[ \text{quotient} = 2(555 - 445) = 220 \]
\[ \text{number} = 1000 \times 220 + 30 = 220{,}030 \]
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Four digits of the number 29138576 are omitted so that the largest result is possible. The largest omitted digit is:
Answer: A
We have to omit four digits in such a manner that we get the largest possible number as the result.
We will omit 2,1,32,1,3 and 55 and the result will be 98769876.
The largest omitted digit will be 5.
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The highest power of 9 dividing 99! completely is
Answer: D
Count the factors of 3 first, then pair them up. Legendre's rule gives the exponent of a prime in a factorial:
\[ \left\lfloor\frac{99}{3}\right\rfloor + \left\lfloor\frac{99}{9}\right\rfloor + \left\lfloor\frac{99}{27}\right\rfloor + \left\lfloor\frac{99}{81}\right\rfloor = 33 + 11 + 3 + 1 = 48 \]
So \( 3^{48} \) divides 99! and no higher power does. Because \( 9 = 3^{2} \), each factor of 9 consumes two of those threes:
\[ \text{highest power of } 9 = \left\lfloor\frac{48}{2}\right\rfloor = 24 \]
Answer 48 is the trap for stopping at the power of 3 without halving it.
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How many of the following numbers are divisible by 132 ?
264, 396, 462, 792, 968, 2178, 5184, 6336
Answer: A
\(132 = 4 \times 3 \times 11\)
So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.
\(264 \Rightarrow 11,3,4\)
\(396 \Rightarrow 11,3,4\)
\(462 \Rightarrow 11,3\)
\(792 \Rightarrow 11,3,4\)
\(968 \Rightarrow 11,4\)
\(2178 \Rightarrow 11,3\)
\(5184 \Rightarrow 3,4\)
\(6336 \Rightarrow 11,3,4\)
Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.
Required number of number = 4.
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The sum of three consecutive even numbers is 42. Find the middle number of the three?
Answer: A
Three consecutive even numbers \((2P - 2), 2P, (2P + 2)\)
\((2P - 2) + 2P + (2P + 2) = 42\)
\(6P = 42 => P = 7\)
The middle number is: \(2P = 14\)
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The unit’s digit in the product \((3127)^{173}\) is:
Answer: C
Unit digit in \((3127)^{173}\) = Unit digit in 7173. Now, 74 gives unit digit 1
Therefore, \(7173= (74)43 \times 71\). Thus, 7173 gives unit digit 7
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