Without performing actual division and without using calculators, find out if 1 0 2 1 9 9 9 5 5 9 is an integer.
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1 0 2 1 0 0 0 1 0 0 0 × 1 0 2 1 − 9 9 9 5 5 9 ⟹ 9 9 9 5 5 9 ⟹ 1 0 2 1 9 9 9 5 5 9 = 2 1 4 4 1 = 2 0 4 2 0 2 0 × 1 0 2 1 + 1 0 2 1 = ( 1 0 0 0 − 2 0 − 1 ) 1 0 2 1 = 1 0 0 0 − 2 0 − 1 = 9 7 9
Yes, 1 0 2 1 9 9 9 5 5 9 is an integer .
1 0 2 1 × 1 0 0 0 = 1 0 2 1 0 0 0 (Trying to get as close as possible to 9 9 9 5 5 9 without getting too accurate)
1 0 2 1 0 0 0 − 9 9 9 5 5 9 = 2 1 4 4 1
1 0 2 1 × 2 0 = 2 0 4 2 0 (Trying to get as close as possible to 2 1 4 4 1 without getting too accurate)
2 1 4 4 1 − 2 0 4 2 0 = 1 0 2 1
2 0 4 2 0 mod 1 0 2 1 ≡ 0 and 1 0 2 1 mod 1 0 2 1 ≡ 0 ⇒ 2 1 4 4 1 mod 1 0 2 1 ≡ 0
1 0 2 1 0 0 0 mod 1 0 2 1 ≡ 0 ⇒ 9 9 9 5 5 9 mod 1 0 2 1 ≡ 0
Therefore, the answer is yes, it is an integer
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9 9 9 5 5 9 = 1 0 0 0 0 0 0 − 4 4 1 = 1 0 0 0 2 − 2 1 2 = ( 1 0 0 0 + 2 1 ) ( 1 0 0 0 − 2 1 ) = 1 0 2 1 × 9 7 9