Assistant Provident Fund Commissioner and Enforcement Officer / Accounts Officer (APFC and EO/AO) 2025 recruitment test. Transcribed from the Commission's own booklet (series A, PENA-A-PTS); answer worked out by our faculty — UPSC has published no answer key for this sitting. This question is in Part B. Answer last reviewed 25 August 2026.

How many digits are there in 6^25 ? (It is given that log10 2 = 0·3010 and log10 3 = 0·4771)

(a)20
(b)21
(c)22
(d)23

Answer

Why

Correct — A, (a) 20. The rule. For a positive number N, the number of digits in N is the integer part of log to base 10 of N, plus one. The reason is that a 20-digit number lies between 10^19 and 10^20, so its logarithm lies between 19 and 20 — the integer part of the logarithm is one less than the digit count, always. The working. log 6^25 = 25 × log 6 log 6 = log (2 × 3) = log 2 + log 3 = 0·3010 + 0·4771 = 0·7781 25 × 0·7781 = 19·4525 Integer part = 19, so the number of digits = 19 + 1 = 20. The multiplication is worth doing carefully, since everything turns on it: 0·7781 × 25 is the same as 0·7781 × 100 ÷ 4 = 77·81 ÷ 4 = 19·4525. Dividing by four is quicker and safer here than multiplying by twenty-five directly. What the fractional part tells you, and what it does not. The 0·4525 fixes the leading digits of the answer, not the number of them; only the integer part 19 counts digits. A candidate who rounds 19·4525 up to 20 and then adds one has used the rule on a rounded value, which is exactly the error the rule is designed to avoid — the rule takes the FLOOR of the logarithm, never the nearest whole number. The two logarithms given in the brackets are the standard pair to memorise: log 2 = 0·3010 and log 3 = 0·4771. From those two, logarithms of 4, 5, 6, 8, 9, 12 and many other small numbers follow — log 5 = 1 − log 2 = 0·6990, log 4 = 2 log 2 = 0·6020, log 9 = 2 log 3 = 0·9542. A second, independent check on the size. 6^25 = (6^5)^5 = 7,776^5, and 7,776 is a little under 8 × 10^3, so the fifth power is a little under 32 × 10^15, that is of the order of 10^19 — a number with twenty digits. Arriving at the same digit count by a route that uses no logarithms at all is the most reassuring check available on this kind of question, and it takes only a moment.

Why the others are wrong

(b)21 — The exact result of rounding before applying the rule: 19·4525 rounded to 20, plus one, gives 21. It is the single commonest error in digit-counting questions. The rule uses the INTEGER PART — the whole number below the logarithm — because that integer is the power of ten the number has just passed. Since 19·4525 lies between 19 and 20, the number lies between 10^19 and 10^20, and every such number has exactly 20 digits.
(c)22 — Arises from arithmetic error in the multiplication — for instance from taking log 6 as about 0·85 or slipping a digit in 0·7781 × 25. Any value of the product between 21 and 22 would give this answer, so the way to guard against it is to sanity-check the size: 6^25 is a little more than 10^19, since log 6 is a little under 0·8 and 25 × 0·8 = 20.
(d)23 — Requires the logarithm to come out above 22, which would mean log 6 was taken as roughly 0·88. That is nearly log 7·6. This option is offered for a candidate who guesses upward from the size of the exponent rather than working with the logarithms supplied, and it is a reminder that the two values in the brackets are given because they are needed.

Concept

The common logarithm of a number splits into two parts: the CHARACTERISTIC, the integer part, which records the power of ten and hence the position of the decimal point, and the MANTISSA, the fractional part, which records the digits. For a number greater than one, the characteristic is one less than the number of digits before the decimal point — which is the whole basis of digit-counting questions. The laws used here are log(ab) = log a + log b and log(a^n) = n log a, which turn a product into a sum and a power into a multiple. That conversion of multiplication into addition is why logarithms were invented, and it is what makes the size of a number like 6^25 accessible without ever computing it.

Digit-counting questions appear regularly in the quantitative block, and they always supply the logarithms needed, so no table has to be remembered beyond log 2 and log 3. The whole item is a test of one rule and one careful multiplication. The examiner's standard distractor is the off-by-one produced by rounding, and it is placed immediately next to the correct answer in the option list. The two logarithms are printed in the stem because they are needed, which is a general feature of this paper's quantitative items: whatever cannot be derived is supplied. That makes the whole question a test of one rule and one multiplication, and the examiner has placed the rounding error immediately next to the correct answer in the option list. When four options are consecutive integers, the arithmetic must be exact and the rule applied to the exact value; there is no scope for estimating, and the neighbouring option is always the trap for whoever estimates.

Key facts

Number of digits in N = (integer part of log N) + 1, for N greater than 1.
log(a^n) = n log a and log(ab) = log a + log b.
log 2 = 0·3010 and log 3 = 0·4771 are the two values worth memorising.
log 6 = log 2 + log 3 = 0·7781.
25 × 0·7781 = 19·4525, so 6^25 lies between 10^19 and 10^20 and has 20 digits.
The characteristic of a logarithm fixes the number of digits; the mantissa fixes the digits themselves.
log 5 = 1 − log 2 = 0·6990, log 4 = 0·6020 and log 9 = 0·9542 all follow from the same two values.
Multiplying by 25 is conveniently done as multiplying by 100 and dividing by 4.

Study next

Laws of logarithms and change of baseCharacteristic and mantissa, including the treatment of numbers less than oneFinding the number of zeros at the end of a factorial, a related counting questionComparing large powers without computing themAntilogarithms and the reconstruction of leading digits

Common traps

Rounding the logarithm before applying the rule, which gives an answer one too large.
Adding one to the mantissa rather than to the characteristic.
Mis-multiplying 0·7781 by 25; use 77·81 ÷ 4 instead.
Forgetting that the rule needs the number to be greater than one.

Logarithm questions in EPFO papers are almost always digit counting or a simple evaluation, and the required log values are printed in the stem. Learn the one rule, practise the multiplication, and treat the neighbouring option as the trap it is. Where the stem supplies constants in brackets, take that as an instruction: the constants are there because the intended solution uses them, and a method that ignores them is probably the wrong method.

Related PYQs

No directly related past PYQ was found.

Practice

practice — not a real PYQ
How many digits are there in 2^30 ? (Take log10 2 = 0·3010)
- (a)9
- (b)10
- (c)11
- (d)12
Answer(b) 10
practice — not a real PYQ
If log10 2 = 0·3010, then log10 5 is equal to which one among the following ?
- (a)0·3010
- (b)0·6020
- (c)0·6990
- (d)0·7781
Answer(c) 0·6990

Why

Why the others are wrong

(b)21 — The exact result of rounding before applying the rule: 19·4525 rounded to 20, plus one, gives 21. It is the single commonest error in digit-counting questions. The rule uses the INTEGER PART — the whole number below the logarithm — because that integer is the power of ten the number has just passed. Since 19·4525 lies between 19 and 20, the number lies between 10^19 and 10^20, and every such number has exactly 20 digits.
(c)22 — Arises from arithmetic error in the multiplication — for instance from taking log 6 as about 0·85 or slipping a digit in 0·7781 × 25. Any value of the product between 21 and 22 would give this answer, so the way to guard against it is to sanity-check the size: 6^25 is a little more than 10^19, since log 6 is a little under 0·8 and 25 × 0·8 = 20.
(d)23 — Requires the logarithm to come out above 22, which would mean log 6 was taken as roughly 0·88. That is nearly log 7·6. This option is offered for a candidate who guesses upward from the size of the exponent rather than working with the logarithms supplied, and it is a reminder that the two values in the brackets are given because they are needed.

Concept

Key facts

Number of digits in N = (integer part of log N) + 1, for N greater than 1.
log(a^n) = n log a and log(ab) = log a + log b.
log 2 = 0·3010 and log 3 = 0·4771 are the two values worth memorising.
log 6 = log 2 + log 3 = 0·7781.
25 × 0·7781 = 19·4525, so 6^25 lies between 10^19 and 10^20 and has 20 digits.
The characteristic of a logarithm fixes the number of digits; the mantissa fixes the digits themselves.
log 5 = 1 − log 2 = 0·6990, log 4 = 0·6020 and log 9 = 0·9542 all follow from the same two values.
Multiplying by 25 is conveniently done as multiplying by 100 and dividing by 4.

Study next

Common traps

Rounding the logarithm before applying the rule, which gives an answer one too large.
Adding one to the mantissa rather than to the characteristic.
Mis-multiplying 0·7781 by 25; use 77·81 ÷ 4 instead.
Forgetting that the rule needs the number to be greater than one.