Let y denote a single-digit integer. The tens digit in the product of 2 y 7 and 3 9 is 9 . Find y .
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tens digit to be * 9 .
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Thanks. Why don't you learn to use LaTex. I have been editing your problems. Members will read your problems and solutions if they are in LaTex.
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I use LaTex already. You just modify my problems slightly in a different format...and that's all ! You know sometimes I feel that I also could modify other's problems when they are badly written...lol.
( 2 0 7 + 1 0 y ) × 3 9 = 8 0 7 3 + 3 9 0 y
The tens digit of this is the units digit of 7 + 9 y . Since we want this to be 9 , we want the units digit of 9 y to be 2 ; this happens when y = 8 .
(200 + 10y + 7)* 39 = 7800 + 390y + 273 = 8073 + 390y as 0<=y<=9 the different possible values of (8073 + 390y) for y =0,1,...9 are 8463,8853,9243,9633,10023,10413, 10803,11193 and 11583. This implies that y = 8
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2 y 7 × 3 9 = 3 9 ( 2 0 7 + 1 0 y ) = 8 0 7 3 + 3 9 0 y
For the tens digit to be 9 , we have 7 + 9 y ≡ 9 (mod 10) . ⟹ 9 y ≡ 2 (mod 10) and y = 8 .