Suppose, the remainder obtained while dividing x by 61 is 2. What is the remainder obtained while dividing x⁷ by 61?

Correct — D, (d) 6. The decisive idea is that a remainder is all that survives multiplication. If two numbers leave the same remainder on division by 61, then their products, their squares and their seventh powers also leave the same remainder. So the value of x is not needed and cannot be found — only its remainder matters, and that remainder is 2. The question therefore becomes: what is the remainder when 2⁷ is divided by 61?

2⁷ = 128, and 128 = 61 × 2 + 6, so the remainder is 6. That is option (d).

Written out, the reason is the binomial expansion. Since x leaves remainder 2 on division by 61, we may write x = 61k + 2 for some whole number k. Raising that to the seventh power gives a long sum in which every single term except the last carries at least one factor of 61, because every term except the last contains 61k. All of those terms are divisible by 61 and contribute nothing to the remainder. The last term is 2⁷ = 128, which is not smaller than 61 and so must itself be reduced: 128 − 122 = 6. Hence x⁷ is 6 more than a multiple of 61.

A candidate who wants certainty can take the smallest permitted value and test it. The number 2 itself leaves remainder 2 on division by 61; 2⁷ is 128; and 128 divided by 61 leaves 6. Any other admissible value — 63, 124, 185 — gives the same answer, which is the whole point of the method. Two habits are worth taking from this item. First, reduce before you multiply, never after: working with 2 rather than with an unknown keeps every number small. Second, reduce again at the end, because a remainder must always be less than the divisor, and 128 is not.

Why the others are wrong

• (a)2 — Two is the remainder of x itself, and this option is chosen by a candidate who assumes that raising a number to a power leaves its remainder untouched. It does not, and the assumption collapses at the very first step: if x leaves remainder 2, then x² leaves remainder 4, not 2, since 2 × 2 = 4 and 4 is already less than 61. What is preserved is the operation, not the value — the remainder of a product is found from the remainders of the factors, but it is not equal to them. There is a special case in which a remainder does survive a power, namely when the base is 0 or 1, or when the exponent happens to bring the cycle of remainders back to its starting value, and it is worth knowing that such cycles exist. Here the cycle of powers of 2 modulo 61 is nowhere near returning at the seventh step: 2, 4, 8, 16, 32, 3, 6.
• (b)4 — Four is 2², so this is the answer to the question the candidate would be solving if the exponent had been 2 rather than 7. It is a copying error rather than a mathematical one, and the small raised 7 in the printed stem is exactly the kind of detail a hurried eye drops — the paper prints x with a superscript seven, in italic, against roman text around it. The check that catches it is to write the exponent down separately before starting: seventh power, base remainder 2, divisor 61. Incidentally, the sequence of remainders left by successive powers of 2 on division by 61 runs 2, 4, 8, 16, 32, 3, 6 for the first through seventh powers, so 4 is genuinely the answer for the second power and 6 for the seventh. Reading the exponent correctly is the whole difference between the two.
• (c)5 — Five is what a slip in the final subtraction produces, and nothing else in the problem yields it. The last step is to reduce 128 below 61, which means subtracting 61 twice: 128 − 61 = 67, and 67 − 61 = 6. A candidate who subtracts 123 instead of 122, or who works from the multiplication table too quickly and takes 61 × 2 as 123, lands on 5. The general safeguard against this class of error is to finish by reconstructing the number: if the remainder is 5 then x⁷ would be 61 × 2 + 5 = 127, and 2⁷ is 128, so the reconstruction fails by one. Rebuilding the original figure from the quotient and the remainder takes a moment and catches every arithmetic slip in the division, which is why it is worth doing on any item where the options are a single digit apart.

This is modular arithmetic, the arithmetic of remainders, and it rests on one property: remainders survive addition and multiplication. If a leaves remainder r on division by n and b leaves remainder s, then a + b leaves the same remainder as r + s, and a × b leaves the same remainder as r × s — in each case after the result is itself reduced below n if necessary. Powers follow immediately, since a power is repeated multiplication, so a⁷ leaves the same remainder as r⁷. This is what makes questions about enormous numbers answerable by hand: the number is replaced by its remainder at the earliest possible moment and never grows again. The formal justification is the binomial expansion of (nk + r)⁷, in which every term but the last contains a factor n and is therefore invisible to the remainder. Two further tools belong to the same topic. The first is cyclicity: the remainders of successive powers of a fixed base eventually repeat, so for a large exponent one finds the length of the cycle and takes the exponent modulo that length. The second, for a prime divisor p and a base not divisible by p, is Fermat's little theorem, which states that the (p − 1)th power leaves remainder 1 — here 61 is prime, so x⁶⁰ would leave remainder 1, a fact of no use at the seventh power but decisive when the exponent runs into the hundreds.

Key facts

• Remainders are preserved under multiplication, so x⁷ leaves the same remainder as 2⁷ when both are divided by 61.
• 2⁷ = 128 and 128 = 61 × 2 + 6, so the required remainder is 6.
• Writing x = 61k + 2 and expanding by the binomial theorem, every term except 2⁷ carries a factor of 61 and cannot affect the remainder.
• A remainder must always be smaller than the divisor, so 128 has to be reduced further even though it is a legitimate intermediate value.
• The powers of 2 leave remainders 2, 4, 8, 16, 32, 3 and 6 on division by 61 for the first through the seventh power.
• For large exponents the remainders repeat in a cycle, and the exponent is reduced modulo the length of that cycle.
• Fermat's little theorem: for a prime p and a base not divisible by p, the (p − 1)th power leaves remainder 1; 61 is prime, so x⁶⁰ would leave remainder 1.
• Any answer can be checked by rebuilding the number: quotient times divisor plus remainder must return the original value.

Common traps

• Assuming a remainder is unchanged by raising to a power; the square of a number leaving remainder 2 leaves remainder 4.
• Misreading the exponent, particularly a small superscript, and answering for the square instead of the seventh power.
• Stopping at 128 without reducing it below the divisor, since a remainder must be smaller than what you divided by.
• Trying to find the value of x, which is neither possible nor necessary — only its remainder is given and only that is needed.
• Slipping in the last subtraction, which on this item moves the answer from 6 to a neighbouring option.

• practice — not a real PYQ

  A number leaves a remainder of 3 when divided by 7. What is the remainder when the cube of that number is divided by 7?

  • (a)1
  • (b)3
  • (c)4
  • (d)6

  Answer(d) 6

• practice — not a real PYQ

  What is the remainder when 2¹⁰ is divided by 11?

  • (a)0
  • (b)1
  • (c)2
  • (d)10

  Answer(b) 1