ACT Math Repeating Patterns: 18 Practice Problems with Step-by-Step Explanations
Repeating pattern problems on the ACT show up in several different costumes — finding a far-off digit in a repeating decimal, figuring out what day of the week it will be hundreds of days from now, simplifying a huge power of i, or finding the ones digit of a number raised to a large exponent — but underneath, every single one of these is the exact same skill: identifying a repeating cycle and using division with a remainder to figure out where in that cycle you land.
The universal technique: find the length of the repeating cycle (5 digits in a repeating decimal like 0.46981, 7 days in a week, 4 for powers of i, or 4 for the units digit of most integers), divide the target position by that cycle length, and use the remainder to find your place in the cycle. The one detail that trips students up: when the remainder is 0, that doesn't mean "nothing" — it means you land exactly on the last position in the cycle, not before the first one. For example, i4, i8, and i12 all equal 1, even though 4, 8, and 12 all divide evenly with no remainder.
Day-of-week problems work the same way using a 7-day cycle, but pay attention to whether the question asks about the future (add days forward) or the past (count backward) — both use the same remainder, just applied in opposite directions. Units-digit problems for powers of integers usually cycle every 4 exponents (though some bases have shorter cycles, like 5 or 6, which always end in 5 or 6 respectively). Work through the 18 problems below, then click to reveal each step-by-step explanation.
On this page:
- Problem 1 — Digit in a repeating decimal
- Problem 2 — Digit in a repeating decimal
- Problem 3 — Power of i
- Problem 4 — Day of the week, forward
- Problem 5 — Digit in a repeating decimal
- Problem 6 — Day of the week, backward
- Problem 7 — Digit in a repeating decimal
- Problem 8 — Combining powers of i
- Problem 9 — Digit in a repeating decimal
- Problem 10 — Units digit of a large power
- Problem 11 — Day of the week, forward
- Problem 12 — Power of i
- Problem 13 — Units digit of a large power
- Problem 14 — Units digit, extending a cycle
- Problem 15 — Power of i with a variable exponent
- Problem 16 — Comparing two powers of i
- Problem 17 — Month cycle
- Problem 18 — Units digit, shifting a cycle
Practice Problems
What is the 81st digit after the decimal point in 0.46981?
- A) 1
- B) 4
- C) 6
- D) 8
The repeating block "46981" has 5 digits. Divide 81 by 5: 81 = 5(16)+1, so the remainder is 1.
A remainder of 1 means the 81st digit matches the 1st digit of the block, which is 4.
What is the 331st digit after the decimal point in the repeating decimal 0.1679?
- A) 0
- B) 1
- C) 3
- D) 7
The repeating block "1679" has 4 digits. Divide 331 by 4: 331 = 4(82)+3, so the remainder is 3.
A remainder of 3 means the 331st digit matches the 3rd digit of the block, which is 7.
i77 is equal to:
- A) −1
- B) −i
- C) i
- D) 1
Powers of i cycle every 4 exponents: i1=i, i2=−1, i3=−i, i4=1, then it repeats. Divide 77 by 4: 77 = 4(19)+1, so the remainder is 1.
A remainder of 1 means i77 = i1 = i.
Given that today is Sunday, what day of the week will it be 132 days from today?
- A) Wednesday
- B) Thursday
- C) Friday
- D) Saturday
The week repeats every 7 days. Divide 132 by 7: 132 = 7(18)+6, so the remainder is 6.
Count forward 6 days from Sunday: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday. So 132 days from Sunday is a Saturday.
What is the 297th digit after the decimal point in the repeating decimal 0.39765?
- A) 5
- B) 6
- C) 7
- D) 9
The repeating block "39765" has 5 digits. Divide 297 by 5: 297 = 5(59)+2, so the remainder is 2.
A remainder of 2 means the 297th digit matches the 2nd digit of the block, which is 9.
Given that today is Thursday, what day of the week was 205 days ago?
- A) Saturday
- B) Sunday
- C) Monday
- D) Tuesday
The week repeats every 7 days. Divide 205 by 7: 205 = 7(29)+2, so the remainder is 2.
Since this is a "days ago" question, count backward 2 days from Thursday: Wednesday, then Tuesday. So 205 days ago was a Tuesday.
What is the 52nd digit after the decimal point in the repeating decimal 0.80345?
- A) 0
- B) 3
- C) 4
- D) 5
The repeating block "80345" has 5 digits. Divide 52 by 5: 52 = 5(10)+2, so the remainder is 2.
A remainder of 2 means the 52nd digit matches the 2nd digit of the block, which is 0.
What is 10i16 − 4i6 equal to?
- A) 4
- B) 6
- C) 10
- D) 14
For i16: 16 divides evenly by 4 with a remainder of 0, which means it lands on the last position in the cycle: i16 = i4 = 1. For i6: 6 = 4(1)+2, so i6 = i2 = −1.
Substitute: 10(1) − 4(−1) = 10 + 4 = 14.
What is the 217th digit after the decimal point in the repeating decimal 0.217657?
- A) 1
- B) 2
- C) 5
- D) 6
The repeating block "217657" has 6 digits. Divide 217 by 6: 217 = 6(36)+1, so the remainder is 1.
A remainder of 1 means the 217th digit matches the 1st digit of the block, which is 2.
When 7217 is multiplied out, what is the digit in the ones place?
- A) 1
- B) 2
- C) 3
- D) 7
The units digit of powers of 7 cycles every 4 exponents: 71=7, 72=...9, 73=...3, 74=...1, then it repeats. Divide 217 by 4: 217 = 4(54)+1, so the remainder is 1.
A remainder of 1 means the units digit matches the 1st position in the cycle, which is 7.
Given that today is Friday, what day of the week will it be in 157 days?
- A) Sunday
- B) Saturday
- C) Monday
- D) Tuesday
Divide 157 by 7: 157 = 7(22)+3, so the remainder is 3.
Count forward 3 days from Friday: Saturday, Sunday, Monday. So 157 days from Friday is a Monday.
What is the value of i263?
- A) i
- B) −1
- C) −i
- D) 1
Divide 263 by 4: 263 = 4(65)+3, so the remainder is 3.
A remainder of 3 means i263 = i3 = −i.
When 3106 is multiplied out, what is the digit in the ones place?
- A) 3
- B) 4
- C) 7
- D) 9
The units digit of powers of 3 cycles every 4 exponents: 31=3, 32=...9, 33=...7, 34=...1, then it repeats. Divide 106 by 4: 106 = 4(26)+2, so the remainder is 2.
A remainder of 2 means the units digit matches the 2nd position in the cycle, which is 9.
If 890 has a 4 in the units digit, what number does 892 have in the units digit?
- A) 0
- B) 2
- C) 4
- D) 6
The units digit of powers of 8 cycles every 4 exponents: 8, 4, 2, 6, then repeats. Since 890 has units digit 4 (the 2nd position in the cycle), advancing 2 more exponents (to 892) moves 2 more positions forward in the cycle: from position 2 to position 4.
The 4th position in the cycle (8, 4, 2, 6) is 6.
If m is an integer, what does i8m−3 equal?
- A) i
- B) −i
- C) −1
- D) 1
Since 8m is always a multiple of 4 (regardless of the integer m), 8m contributes a remainder of 0 when divided by 4. So the exponent 8m−3 behaves the same as −3 in terms of its position in the cycle.
−3 mod 4 = 1 (since −3+4=1), so i8m−3 = i1 = i for every integer m.
i102/i80 = ?
- A) i12
- B) i13
- C) i14
- D) i15
Dividing powers with the same base subtracts the exponents: i102/i80 = i102−80 = i22. Since 22 = 4(5)+2, i22 = i2 = −1.
Check each answer choice for the one that also simplifies to −1: i14, since 14 = 4(3)+2, also gives i2 = −1. That's the match.
Barney purchased a new laptop in June that is estimated to stop working in 90 months. What month is his laptop estimated to stop working?
- A) February
- B) March
- C) October
- D) December
The year repeats every 12 months. Divide 90 by 12: 90 = 12(7)+6, so the remainder is 6.
Count forward 6 months from June: July, August, September, October, November, December. So 90 months from June is December.
If 7x−2 has a 3 in the ones place, what digit does 7x+3 have in the ones place?
- A) 1
- B) 3
- C) 7
- D) 9
A units digit of 3 for a power of 7 corresponds to the 3rd position in its 4-term cycle (7, 9, 3, 1). The exponent increases from (x−2) to (x+3), a jump of 5. Since the cycle length is 4, a jump of 5 is equivalent to a jump of 1 position (5 mod 4 = 1).
Moving 1 position forward from the 3rd position (3) lands on the 4th position, which is 1.
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