EPFO EO/AO 2023 General Ability Test · Question 73 of 120

Correct — C, (c) 6. The booklet prints this as a power tower — a base 2 carrying the superscript 2, which itself carries the superscript 100 — and it is reproduced here as 2^(2^100) because JSON has no superscripts. Nothing in the question is to be computed. 2^100 alone is a thirty-one digit number, and 2^(2^100) has roughly 3.8 × 10²⁹ digits. The whole item is about cyclicity. Step one: the unit digits of the powers of 2 repeat with period 4. 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, 2⁶ = 64, 2⁷ = 128, 2⁸ = 256 — the last digits run 2, 4, 8, 6, 2, 4, 8, 6 and never do anything else. So the unit digit of 2^N depends on nothing about N except its remainder on division by 4: remainder 1 gives 2, remainder 2 gives 4, remainder 3 gives 8, and remainder 0 gives 6. Step two: find the exponent's remainder modulo 4. The exponent here is N = 2^100. Since 100 is at least 2, 2^100 = 4 × 2^98, which is plainly a multiple of 4. So N ≡ 0 (mod 4). Step three: remainder 0 corresponds to the fourth entry of the cycle, which is 6. Hence 2^(2^100) ends in 6. Notice how much this argument rules out before it ever reaches an answer. The exponent is itself a power of 2 and therefore even, so its remainder modulo 4 can only be 0 or 2 — never 1 and never 3. That means the unit digit of a tower of this shape can only ever be 6 or 4, and options (a) and (d) are impossible before any calculation is done. The remaining work is only to decide between 4 and 6, and 2^100 being divisible by 4 settles it. One word on reading the tower. A stack of exponents is evaluated from the top down: 2^(2^100), not (2^2)^100. The two are different numbers — the second is only 2^200 — and confusing them is a real error, even though on this particular question it does not change the answer, since 200 is also a multiple of 4 and also lands on 6.

• (a)2 — A unit digit of 2 requires the exponent to leave remainder 1 on division by 4, that is, to be odd. The exponent here is 2^100, which is even, so this can never happen for a tower of this shape. The usual route to this answer is an off-by-one in the cycle: having correctly found that the exponent is a multiple of 4, the candidate reads 'remainder 0' as 'the start of the cycle' and picks the cycle's first entry, 2. Remainder 0 means the last entry, not the first — the fourth power, 16, ends in 6, and it is 2^1 that ends in 2.
• (b)4 — This is the answer if the exponent leaves remainder 2 on division by 4, so unlike options (a) and (d) it is at least of the right parity and cannot be dismissed on sight. The specific error that produces it is reducing the exponent by its own last digit instead of by its remainder modulo 4: 2^100 does end in 6, and 6 leaves remainder 2 on division by 4, which points at 4. But the cycle is driven by the exponent modulo 4, not by the exponent's unit digit, and 2^100 is divisible by 4 outright.
• (d)8 — A unit digit of 8 requires the exponent to leave remainder 3 on division by 4, which means an odd exponent. The exponent 2^100 is even, so 8 is impossible here for the same structural reason as option (a). It is offered because 8 is the third entry of the cycle 2, 4, 8, 6, and a candidate who miscounts the position by one in the other direction lands on it. Fixing the correspondence once — remainders 1, 2, 3, 0 map to 2, 4, 8, 6 — closes both of these traps at once.

The last digit of a power depends only on the last digit of the base and on the exponent's remainder in a short cycle. For base 2 the cycle is 2, 4, 8, 6 with period 4; for 3 it is 3, 9, 7, 1; for 7 it is 7, 9, 3, 1; for 8 it is 8, 4, 2, 6 — all of period 4. Bases ending in 4 and 9 have period 2 (4, 6 and 9, 1), and bases ending in 0, 1, 5 or 6 have period 1, always ending in themselves. The working rule is: divide the exponent by the period and use the remainder, treating remainder 0 as the last entry of the cycle. When the exponent is itself an enormous number, the only question is what that number's remainder is, and that is usually answerable by a divisibility observation rather than by computing it. Here the exponent is 2^100 and the only fact needed about it is that 4 divides it. The same technique answers questions about the last two digits (using cycles modulo 100), about remainders of large powers (using Fermat's little theorem or Euler's theorem), and about which of two towers is larger. Whenever an exam prints a tower, it is signalling that no arithmetic is intended.

The EO/AO quantitative block includes one or two items each sitting that look impossible and take fifteen seconds once the method is known, and this is one of them. Their function is to reward the candidate who recognises a standard technique instead of reaching for a calculation. The habit rewarded is to identify what actually determines the answer — here, the exponent's remainder modulo 4 — and then to look for the cheapest way of getting that one fact.

• The unit digits of the powers of 2 cycle with period 4: 2, 4, 8, 6, and then repeat.
• The unit digit of 2^N is determined by N mod 4 — remainder 1 gives 2, remainder 2 gives 4, remainder 3 gives 8, remainder 0 gives 6.
• 2^100 is divisible by 4, since 2^100 = 4 × 2^98, so the exponent leaves remainder 0 and the unit digit is 6.
• Because the exponent 2^100 is itself even, the unit digit of 2^(2^100) can only be 4 or 6; 2 and 8 are impossible for a tower of this shape.
• A power tower is evaluated from the top downwards: 2^(2^100) is not the same number as (2^2)^100 = 2^200.
• 2^100 is a thirty-one digit number, and 2^(2^100) has about 3.8 × 10²⁹ digits, so no part of this question is meant to be computed.
• Unit-digit cycles of the other bases: 3 gives 3, 9, 7, 1; 7 gives 7, 9, 3, 1; 8 gives 8, 4, 2, 6; 4 gives 4, 6; 9 gives 9, 1; and 0, 1, 5, 6 repeat themselves.

• Trying to evaluate the tower, or even the exponent. Neither number is writable.
• Reducing the exponent by its own unit digit rather than by its remainder modulo 4.
• Mapping remainder 0 to the first entry of the cycle. Remainder 0 corresponds to the fourth power, which ends in 6.
• Reading 2^(2^100) as (2^2)^100. Towers are evaluated downwards from the top; here both readings happen to end in 6, but that is a coincidence of this question.

Unit-digit and remainder questions appear in almost every EO/AO quantitative block, usually with an exponent chosen so that the exponent's remainder is itself a small puzzle. Learn the four-length cycles for 2, 3, 7 and 8, the two-length cycles for 4 and 9, and the rule for remainder 0, and every question of this type reduces to one division.

No directly related past PYQ was found.

• practice — not a real PYQ

  What is the digit in the unit place of 7^123 ?

  • (a)1
  • (b)3
  • (c)7
  • (d)9

  Answer(b) 3

• practice — not a real PYQ

  The unit digit of 3^(4^50) is :

  • (a)1
  • (b)3
  • (c)7
  • (d)9

  Answer(a) 1