Competition · AMC preparation · step 4 of 4
AMC 8 · 2007 · #18
Grade 5 number-theoryPick an answer.
AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #16 (Change Focus) is the key move: instead of computing the whole 198-digit product, focus only on what the question actually needs — the last four digits. Standard column multiplication shows that the last four digits of a product depend only on the last four digits of each factor; everything to the left feeds into higher places and never comes back down. Tool #9 (Easier Problem) confirms the shortcut by trying a baby version (much shorter 303… and 505… numbers) and checking that the last four digits of the product are unchanged.
Focus on the last four digits
Ignore the whole product: the thousands and units digits sit in its last four places, so only each factor's last four digits reach them.
Place-value thinking from Grade 5: a digit in the ten-thousands place or higher cannot land in the ones, tens, hundreds, or thousands column of the answer.
The last four digits of the product are fixed by only the last four digits of each factor, so every digit further to the left can be ignored.
▸ Why?
Each factor equals its last four digits plus a whole number of ten-thousands, and multiplying two such sums distributes into four partial products — one of them being the product of the two last-four-digit parts.
▸ Why?
The three partial products that each carry a factor of ten-thousand end in four zeros, so they add nothing to the ones, tens, hundreds, and thousands columns, leaving only the product of the two last-four-digit parts to set those columns.
▸ Why?
Multiplying a whole number by ten-thousand shifts every digit four places higher, filling the ones through thousands places with zeros.
▸ Why?
Adding a number that is zero in those four places leaves each of those columns exactly as it already was.
Read off each factor's ending
Read each factor's last four digits: N₁ ends 0303, N₂ ends 0505 — so we multiply 303 by 505.
Reading the rightmost four digits of a multi-digit number is exactly Grade 5 place-value identification.
5.NBT.A.1Change Focus Count The ComplementMultiply the two small numbers
Multiply the two small numbers by partial products to get 153015.
Replacing two 99-digit numbers with two 3-digit numbers is the Easier Problem move. The product 303 × 505 is a clean Grade 5 multi-digit multiplication.
5.NBT.B.5Solve An Easier Related ProblemTake the last four digits
Its last four digits are 3015, so the thousands digit A = 3, the units digit B = 5, and A + B = 8.
Picking out the thousands digit and the units digit from a written numeral is the Grade 5 place-value definition in action.
5.NBT.A.1Change Focus Count The ComplementWhen you only need the last few digits of a giant product, throw away every digit to the left — Grade 5 place value reduces this AMC 8 problem to a single tidy multiplication.
- Focus on the last four digits
- Read off each factor's ending
- Multiply the two small numbers
- Take the last four digits
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