A number when divided by 6 leaves a remainder 3. When the square of the same number is divided by 6, the remainder is:
Answer: C
Let \(x=6q+3\). Then,
\(x^{2} = (6q+3)^{2} \\= 36q^{2}+36q+9 \\= 6(6q^{2}+6q+1)+3\)
So, when \(2n\) is divided by \(4\), remainder \(=3\)
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The unit digit in \(7^{153}\) is?
Answer: C
\(7^{153} = \left ( 7^{4}\right )^{38}\times 7\)
Now, unit digit of \(\left ( 7^{4}\right )^{38} = 1\)
Therefore, unit digit of \(7153 = 1 \times 7\)
\(= 7\)
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When a number divided by 342 gives remainder 47. When the same number is divided by 19, what would be the remainder?
Answer: B
On dividing the given number by 342, let kk be the quotient and 47 as remainder.
Then,
number−342k+47=(19×18k+19×2+9)number−342k+47=(19×18k+19×2+9)
=19(18k+2)+9=19(18k+2)+9
⇒ The given number when divided by 19, gives 18k+218k+2 as quotient and 9 as remainder.
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It is being given that \((2^{32} + 1)\) is completely divisible by a whole number. Which of the following numbers is completely divisible by this number?
Answer: D
Let \(2^{32} = x\).
Then, \((2^{32} + 1) = (x + 1)\).
Let \((x + 1)\) be completely divisible by the natural number N. Then,
\((2^{96} + 1) \\= [(2^{32})^{3} + 1]\\= (x^{3} + 1)\\= (x + 1)(x^{2} - x + 1)\)
which is completely divisible by N, since \((x + 1)\) is divisible by N.
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\(14927 \times 567 - 14927\times 467 = y\) . What is \(y\) ?
Answer: B
\(y = 14927 \times 567 - 14927 \times 467\)
\(= 14927 \times (567 - 467)\)
\(= 14927 \times 100\)
\(= 1492700\)
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If a set of toys is arranged in rows of 20 each, 15 toys are left over; in rows of 25 each, 20 are left over; in rows of 38 each, 33 are left over; and in rows of 40 each, 35 are left over. What is the minimum number of toys?
Answer: D
In every case the shortfall is the same:
\[ 20 - 15 = 25 - 20 = 38 - 33 = 40 - 35 = 5 \]
So \( N + 5 \) is divisible by all four numbers.
\[ \text{LCM}(20, 25, 38, 40) = 2^{3} \times 5^{2} \times 19 = 3800 \]
\[ N = 3800 - 5 = 3795 \]
Spotting the common shortfall is what turns four separate congruences into a single LCM.
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How many keystrokes are needed to type numbers from 1 to 1000 on a standard keyboard?
Answer: B
While typing numbers from 1 to 1000, you have 9 single digit numbers from 1 to 9. Each of them requires one keystroke. That is 9 key strokes.
There are 90 two-digit numbers, from 10 to 99. Each of these numbers requires 2 keystrokes. Therefore, one requires 180 keystrokes to type the 2 digit numbers.
There are 900 three-digit numbers, from 100 to 999. Each of these numbers requires 3 keystrokes. Therefore, one requires 2700 keystrokes to type these 3 digit numbers.
Then 1000 is a four-digit number which requires 4 keystrokes.
Totally, therefore, one requires 9+180+2700+4=9+180+2700+4= 2893 keystrokes.
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What is the unit digit of \(13^{35}\) ?
Answer: D
\(13^{1}\) has unit digit \(3\)
\(13^{2}\) has unit digit \(9\)
\(13^{3}\) has unit digit \(7\)
\(13^{4}\) has unit digit \(1\)
This will form a pattern \(3-7-9-1\)
So, \(13^{32}\) has unit digit \(1\)
\(13^{35}\) has unit digit as \(7\)
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When a number is divided by 17, the remainder is 9. When the same number is divided by 23, the remainder is 4. Find the number?
Answer: B
\(x = 17p + 9\) and \(x = 23q + 4\)
i.e. \(17p + 9 = 23q + 4\)
Therefore, \(q = \frac{(17p + 5)}{23}\)
Least value of p for which q is a whole number is \(p = 20\)
\(x = 17p + 9\)
\(= 17 \times 20 + 9\)
\(= 349\)
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The sum of the two digits of a number is 10. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number?
Answer: B
Any two digit number can be written as \((10P + Q)\), where \(P\) is the digit in the tens place and \(Q\) is the digit in the units place.
\(P + Q = 10\) ----- (1)
\((10Q + P) - (10P + Q) = 54\)
\(9(Q - P) = 54\)
\((Q - P) = 6\) ----- (2)
Solve (1) and (2) \(P = 2\) and \(Q = 8\)
The required number is \(= 28 \)
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