A and B are playing mathematical puzzles. A asks B "the whole numbers, greater than one, which can divide all the nine three digit numbers 111,222,333,444,555,666, 777,888 and 999?"
B immediately gave the desired answer. It was:
Answer: B
Each of the number can be written as a multiple of 111.
The factors of 111 are 3 and 37
Thus the desired answer is 3, 37 and 111
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Which one of the following is not a prime number?
Answer: D
91 is divisible by 7. So, it is not a prime number.
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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
Required number,
\(= (555+445)\times 2\times 110+30\\ = 220000+30\\ =220030\)
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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
No answer description available for this question.
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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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