In writing all the integers from 1 to 300, how many times is the digit 1 used ?

Correct — A, (a) 160.

THE STEP THAT DECIDES THIS QUESTION is to count each PLACE VALUE separately — units, tens, hundreds — instead of trying to work through the numbers one by one. A digit can appear more than once in the same number, and 111 must be counted three times, so the object being counted is an OCCURRENCE of the digit, not a number containing it.

THE UNITS PLACE. A number ends in 1 once in every run of ten: 1, 11, 21, 31, and so on up to 291. From 1 to 300 there are thirty complete runs of ten, so the digit 1 appears in the units place THIRTY times.

THE TENS PLACE. A number has 1 in its tens place for ten consecutive numbers in every hundred: 10 to 19, then 110 to 119, then 210 to 219. Three such blocks of ten fall below 300, so the digit 1 appears in the tens place THIRTY times.

THE HUNDREDS PLACE. A number has 1 in its hundreds place throughout 100 to 199 and nowhere else in this range. That is a solid block of ONE HUNDRED occurrences, and it is where most of the answer lives.

30 + 30 + 100 = 160

and the answer is option (a). The number 300 itself contributes nothing, so whether the range is read as ending at 299 or at 300 makes no difference.

A CLEANER WAY TO SEE THE SAME COUNT. Write every number from 000 to 299 as a three-digit string, padding with zeros — 001, 002, and so on — which gives exactly three hundred strings, one for each number from 1 to 300 apart from 300 itself, which contains no 1 anyway. Now look down each column.

The units column runs 0–9 repeatedly, so a 1 appears in one tenth of 300 strings: 30 times. The tens column also cycles through 0–9 in blocks, so again 30 times. The hundreds column takes the value 1 for the whole block 100–199: 100 times.

Padding with zeros is a useful device precisely because it makes every position behave uniformly, and the uniform positions can then be counted by simple proportion.

THE DISTINCTION THAT THE QUESTION IS REALLY ABOUT. "How many times is the digit 1 USED" is not the same question as "how many numbers contain the digit 1". The first counts occurrences and gives 160. The second counts numbers and gives a smaller figure, because 11 is one number using the digit twice and 111 is one number using it three times. If it were asked, the second would be answered by counting the numbers with NO 1 in them — for the hundreds digit, two choices out of 0 and 2 when the number is written as a three-digit string; for each of the other two digits, nine choices out of ten — which gives 2 × 9 × 9 = 162 strings, hence 162 numbers from 1 to 300 with no 1 at all, and 300 − 162 = 138 numbers that contain one. A hundred and sixty occurrences spread over a hundred and thirty-eight numbers: the gap is entirely due to numbers such as 11, 110, 121 and 111 which use the digit more than once.

Reading the ask precisely is what this item tests. The stem says "how many times is the digit 1 used", and each written 1 counts.

Why the others are wrong

• (b)140 — One hundred and forty is exactly the count for the range 1 to 200 rather than 1 to 300, which makes it the answer of a candidate who loses the last hundred numbers. Check it: from 1 to 200 the digit 1 appears twenty times in the units place (1, 11, 21, … , 191), twenty times in the tens place (10 to 19 and 110 to 119) and one hundred times in the hundreds place (the block 100 to 199), giving 20 + 20 + 100 = 140. The missing contributions are the ten occurrences in 201 to 300's units places — 201, 211, 221, … , 291 — and the ten in the block 210 to 219, twenty occurrences in all. The slip is easy to make because the hundreds block ends at 199 and a reader who has just finished counting a large block naturally feels the work is done. The remedy is mechanical: settle the range first, write down the three place-value counts as three separate numbers, and only then add them.
• (c)120 — One hundred and twenty is the number of times the digit 1 is used in writing the numbers from 100 to 199 alone, and nothing else. Within that block the hundreds digit is 1 in all hundred numbers, the tens digit is 1 in the ten numbers 110 to 119, and the units digit is 1 in the ten numbers 101, 111, 121, … , 191, so 100 + 10 + 10 = 120. It is the count of a candidate who correctly identifies where most of the ones live, works that block out thoroughly, and then forgets that the digit also appears outside it. Everything below 100 and everything from 200 upwards has been discarded: the twenty occurrences at 1, 11, 21, … , 91 and 10 to 19, and the twenty at 201, 211, … , 291 and 210 to 219. Forty occurrences lost. Counting by place value across the WHOLE range, rather than by block, prevents this, because each place-value count automatically sweeps the entire range.
• (d)110 — One hundred and ten is the smallest option and corresponds to the two most conspicuous groups only — the hundred numbers from 100 to 199, where the hundreds digit is 1, plus the ten numbers from 10 to 19, where the tens digit is 1. It is the count of a candidate who spots the big block, adds the first small run he notices, and stops. Everything in the units place is missing, which is thirty occurrences on its own, and two of the three tens-place runs are missing as well. There is a quick sanity check that rejects it without any recounting: the block 100 to 199 alone contributes at least a hundred occurrences from its hundreds digits, and it must contribute more than that, because some of those numbers also carry a 1 in their tens or units place. So any answer that exceeds one hundred by only ten has plainly not counted the ones INSIDE the hundreds block, let alone those outside it.

COUNTING BY PLACE VALUE is the technique this question exists to teach, and it applies to every problem of the form "how many times does digit d appear in writing the numbers from 1 to N".

The principle is that the three positions of a three-digit number are INDEPENDENT. Whether a number has a 1 in its units place has nothing to do with whether it has a 1 in its tens place, so the total number of ones written is simply the number written in the units column plus the number written in the tens column plus the number written in the hundreds column. Attempting the count number by number, deciding for each of three hundred numbers how many ones it carries, is not only slower but far more error-prone, because the numbers that carry two or three are exactly the ones a hurried counter treats as carrying one.

THE PADDING TRICK makes the columns uniform. Write every number in the range with a fixed number of digits, using leading zeros: 001, 002, … , 299. Now each column cycles regularly. Over any block of ten consecutive strings the units column takes each digit exactly once; over any block of a hundred the tens column takes each digit exactly ten times; and so on. Counting then becomes proportional arithmetic, and it works for any digit — except that the digit ZERO needs care, because leading zeros are not actually written.

THE GENERAL SHAPE, worth carrying. For a complete range 1 to 10^k − 1, each of the nine non-zero digits appears equally often, and the total number of digits written can be found by adding one digit for each of the nine one-digit numbers, two for each of the ninety two-digit numbers, three for each of the nine hundred three-digit numbers, and so on. That companion calculation — how many DIGITS are needed to write out a range — is the other standard question in this family, and it uses the same block-by-block reasoning.

OCCURRENCES AGAINST OBJECTS is the distinction that must be settled before any counting begins. "How many times is the digit used" counts occurrences. "How many numbers contain the digit" counts numbers, and is usually easiest by complement: count the numbers that avoid the digit entirely, using the multiplication principle over the digit positions, and subtract from the total. The two questions have different answers whenever a digit can repeat within one number, which is nearly always, and a single careless reading of the stem sends the whole solution down the wrong road.

Key facts

• The digit 1 appears thirty times in the units place from 1 to 300 — once in every run of ten, at 1, 11, 21, … , 291.
• It appears thirty times in the tens place, in the three blocks 10 to 19, 110 to 119 and 210 to 219.
• It appears one hundred times in the hundreds place, throughout the block 100 to 199, which is where most of the total lies.
• The total is 30 + 30 + 100 = 160 occurrences, and the number 300 itself contributes none, so the endpoint does not affect the count.
• The question counts OCCURRENCES, not numbers: 11 supplies two ones and 111 supplies three, all of which are counted separately.
• Only 138 of the numbers from 1 to 300 contain the digit 1 at all, because 300 − 162 numbers avoid it, so 160 occurrences are spread over 138 numbers.
• Padding every number to three digits with leading zeros makes each column cycle uniformly, so a given digit occupies each column in one tenth of the strings.
• Every wrong option on this item is an incomplete count — 140 is the total for 1 to 200, and 120 is the total for the block 100 to 199 alone.

Common traps

• Counting numbers that contain a 1 rather than the times the digit 1 is written. The two answers differ because 11 and 111 use it more than once.
• Stopping at 200 after finishing the block 100 to 199, which loses twenty occurrences and gives 140.
• Counting only inside the hundreds block and forgetting the ones below 100 and above 199, which gives 120.
• Forgetting the units place altogether, which is worth thirty occurrences on its own and is the easiest column to overlook.
• Working number by number instead of column by column, which is slow and mishandles precisely the numbers that carry the digit twice.

• practice — not a real PYQ

  In writing all the integers from 1 to 100, how many times is the digit 9 used altogether ?

  • (a)10
  • (b)11
  • (c)19
  • (d)20

  Answer(d) 20 — the digit 9 stands in the units place ten times, at 9, 19, 29, … , 99, and in the tens place ten times, throughout 90 to 99. The number 99 is counted twice because the question asks how many times the digit is used, not how many numbers contain it.

• practice — not a real PYQ

  In writing all the integers from 1 to 300, how many of those integers contain the digit 1 at least once ?

  • (a)130
  • (b)138
  • (c)160
  • (d)162

  Answer(b) 138 — count by complement. Written as three-digit strings, a number avoiding the digit 1 has two choices for its hundreds digit, nine for its tens and nine for its units, giving 2 × 9 × 9 = 162 such numbers up to 300, and 300 − 162 = 138 contain a 1. The figure 160 counts occurrences instead of numbers.