A number theory problem.

Number Theory Level 2

1978 ⋅ 1979 + 1980 ⋅ 21 + 1958 1980 ⋅ 1979 − 1978 ⋅ 1979 = ? \large \frac {1978 \cdot 1979 + 1980 \cdot 21 + 1958}{1980 \cdot 1979 - 1978 \cdot 1979 } = ?

Solve the above without using a calculator.


The answer is 1000.

Nguyễn Hữu Khánh by Nguyễn Hữu Khánh , 25, Vietnam

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1 solution

Chew-Seong Cheong
Aug 15, 2017

X = 1978 ⋅ 1979 + 1980 ⋅ 21 + 1958 1980 ⋅ 1979 − 1978 ⋅ 1979 Let a = 1980 = ( a − 2 ) ( a − 1 ) + 21 a + a − 22 a ( a − 1 ) − ( a − 2 ) ( a − 1 ) = a 2 − 3 a + 2 + 21 a + a − 22 a 2 − a − ( a 2 − 3 a + 2 ) = a 2 + 19 a − 20 2 a − 2 = ( a − 1 ) ( a + 20 ) 2 ( a − 1 ) = a + 20 2 = 1980 + 20 2 = 1000 \begin{aligned} X & = \frac {1978 \cdot 1979 + 1980 \cdot 21 + 1958}{1980 \cdot 1979 - 1978 \cdot 1979} & \small \color{#3D99F6} \text{Let } a = 1980 \\ & = \frac {(a-2)(a-1)+21a + a-22}{a(a-1)-(a-2)(a-1)} \\ & = \frac {a^2-3a+2+21a + a-22}{a^2-a-(a^2-3a+2)} \\ & = \frac {a^2+19a -20}{2a-2} \\ & = \frac {(a-1)(a+20)}{2(a-1)} \\ & = \frac {a+20}2 = \frac {1980+20}2 = \boxed{1000} \end{aligned}

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