Two problems. (1) The sequence repeats the block 1, 2, 4, 6, 7, 1. (2) Powers of 8 cycle their ones digit with length 4. Positions and counts are whole numbers. Find the 100th term, how many 1s appear up to the 247th, and the remainder of 8 to the 97th when divided by 5.
Your answer
Watch outIt looks like one 1 per block, but this block carries a 1 at each end — two of them.
How to solve
StrategyLook for a Pattern — Both parts are repeating cycles. For the sequence, the block (1,2,4,6,7,1) of length 6 repeats, so dividing the position by 6 and reading the remainder locates any term and counts the 1s. For the power, the units digit of 8 multiplied repeatedly cycles 8, 4, 2, 6 with length 4, and the remainder mod 5 follows from the units digit. Trying a few small powers reveals that cycle.
1STEP 1
P1: identify the repeating block
The block has length 6 and holds two 1s.
block = (1, 2, 4, 6, 7, 1), length 6
Spotting the block length 6 means every 6 steps the same numbers return.
4.OA.C.5Look For A Pattern
2STEP 2
P1(1): find the 100th number
100 leaves 4, so the 100th term is 6.
100 ÷ 6 = 16 remainder 4; block position 4 = 6
After 16 whole blocks we are 4 steps into the next block, which lands on 6.
4.NBT.B.6Look For A Pattern
3STEP 3
P1(2): count the digit 1 up to position 247
247 is 41 blocks plus one term, giving 82 + 1 = 83 ones.
Only the last digit affects the next last digit, so the units digits must repeat in a short cycle.
3.OA.D.9Solve An Easier Related Problem
Only the last digit affects the next last digit, so the units digits of the powers must repeat in a short cycle.
Why?
Everything above the ones place only ever contributes whole tens to a product, so it can never change the ones digit.
🧱Place-value groupingTen ones make one ten and ten tens make one hundred — each place bundles ten of the last.
Why?
With only ten possible last digits, a repeat must come sooner or later, and once one repeats the whole run of digits repeats after it.
🧱A cycle returns after a full periodA fixed repeating cycle of length p comes back to the same position every p steps.
5STEP 5
P2: locate 8 to the 97th power in the cycle
97 leaves 1, so the ones digit is 8.
97 ÷ 4 = 24 remainder 1; units digit = 8
After 24 full cycles, the 97th power sits at cycle position 1, whose units digit is 8.
4.NBT.B.6Look For A Pattern
6STEP 6
P2: get the remainder mod 5
A ones digit of 8 leaves 3 when divided by 5.
… 8 ÷ 5: 8 = 5 + 3 → remainder 3
Dividing by 5 only cares about the ones digit, and a number ending in 8 leaves remainder 3.
4.NBT.B.6Look For A Pattern
Answer
6, 83 times, remainder 3
P1(1): position 100 with remainder 4 lands on the 4th block entry, 6 -- and indeed positions 4, 10, 16, ... are all 6. P1(2): 41 blocks cover 246 terms (41 x 6), and the 247th term is the next block's first 1, so 82 + 1 = 83 is right. P2: spot-check 8² = 64, 64 / 5 = 12 r 4, and units digit 4 indeed gives remainder 4; our 97th case ends in 8, giving remainder 3, consistent.
Takeaway
When numbers repeat in a loop, just divide the position by the loop length -- the remainder tells you exactly where you land, no need to write them all out!