x divides \(\left ( \left ( 49 \right )^{15} -1\right )\) completely. What is \(x\) ?
Answer: A
\(\left (x^{n}-1 \right )\) is divisible by \(x+1\) if \(x\) is an even number.
\(\left ( \left ( 49 \right )^{15} -1\right )\) =\(\left ( \left ( 7^{2} \right )^{15} -1\right )\) =\(\left ( 7 \right )^{30} -1\)
which is divisible by \(7+1\), \(8\)
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The sum of three consecutive integers is 102. Find the lowest of the three?
Answer: D
Three consecutive numbers can be taken as \((P - 1), P, (P + 1)\)
So, \((P - 1) + P + (P + 1) = 102\)
\(3P = 102 => P = 34\)
The lowest of the three \(= (P - 1) = 34 - 1 = 33\)
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\((112 \times 5^{4}) = ?\)
Answer: B
\[ 5^{4} = 625 \]
\[ 112 \times 625 = 112 \times \frac{10000}{16} = \frac{1{,}120{,}000}{16} = 70{,}000 \]
Writing 625 as \( \tfrac{10^{4}}{16} \) turns the multiplication into a division by 16.
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\((935421 \times 625) = ?\)
Answer: B
Write 625 as \( \tfrac{10^{4}}{16} \).
\[ 935421 \times 625 = \frac{9{,}354{,}210{,}000}{16} = 584{,}638{,}125 \]
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When writing numbers from 1 to 10,000, how many times is the digit 9 written ?
Answer: C
Treat every number from 0000 to 9999 as a four-digit string with leading zeros. That is 10,000 strings and 40,000 digit positions, each digit equally likely:
\[ \frac{40{,}000}{10} = 4000 \]
The padding adds only zeros, so the count of 9s is unaffected, and 10,000 itself contains no 9. So the digit 9 is written 4000 times.
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The numerator of a certain fraction is 8 less than the denominator. If 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes \(\dfrac{3}{4}\). Find the original fraction?
Answer: C
The denominator be \(P\), the numerator will be \((P - 8)\)
The fraction will be \(\dfrac{(P - 8)}{P}\)
Adding 3 to the numerator and subtracting 3 from the denominator,
\(\dfrac{(P - 8 + 3)}{P - 3}= \dfrac{3}{4}\)
\(\dfrac{(P - 5)}{P - 3}= \dfrac{3}{4}\)
\(P = 20 - 9 \Rightarrow P = 11\)
The fraction is: \(\dfrac{3}{11}\)
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a,b,c,d and e are five consecutive numbers in increasing order of size. Deleting one number from the set decreased the sum of the remaining numbers in the set by 20%.
Which one of the following numbers was deleted?
Answer: C
Five consecutive numbers are symmetric about the middle one, so their total is five times that middle value. Writing the set as \( c-2,\, c-1,\, c,\, c+1,\, c+2 \):
\[ S = 5c \]
Removing one member cuts the total by 20 per cent, so the number removed is
\[ 0.20 \times 5c = c \]
which is the middle number, c.
The result holds whatever the five numbers are, since the arithmetic never uses their actual values - only that they are consecutive and so centred on \( c \).
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What least number must be added to 1056, so that the sum is completely divisible by 23 ?
Answer: A
Divide and look at the remainder.
\[ 1056 = 23 \times 45 + 21 \]
\[ 23 - 21 = 2 \]
So adding 2 gives \( 1058 = 23 \times 46 \). Adding the remainder itself would make it divisible only by coincidence - it is the shortfall to the next multiple that is wanted.
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The remainder when the positive integer m is divided by 7 is x. The remainder when m is divided by 14 is x+7.
Which one of the following could m equal?
Answer: B
Test each candidate against both conditions, since the second is far more restrictive than it looks:
| m | m mod 7 = x | m mod 14 | x + 7 | fits? |
|---|---|---|---|---|
| 45 | 3 | 3 | 10 | no |
| 53 | 4 | 11 | 11 | yes |
| 72 | 2 | 2 | 9 | no |
| 85 | 1 | 1 | 8 | no |
The pattern behind it: dividing by 14 leaves either \( x \) or \( x+7 \), and the second happens exactly when the quotient by 7 is odd. Here \( 53 = 7 \times 7 + 4 \), and that quotient 7 is odd, so the remainder on 14 is \( 4 + 7 = 11 \).
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The difference between the local value and face value of 7 in the numeral 657903 is:
Answer: B
The local (place) value of a digit is the digit multiplied by its place; the face value is the digit itself.
In 657903 the 7 sits in the thousands place:
\[ 7000 - 7 = 6993 \]
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