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標題:
functions of equations
發問:
a two- digit number is decreased by 54 when its digits are reversed .if the sum of the squares of the digits is 50, find the number
最佳解答:
Let the two digit number be 10x+y then, we have x2 + y2 = 50 ---- (1) and 10y + x = ( 10x + y ) - 54 that is, 9x - 9y = 54 x - y = 6 ---- (2) sub x = y + 6 into (1) gives (y+6)2 + y2 = 50 2y2 + 12y + 36 - 50 = 0 y2 + 6y - 7 = 0 y = 1 or -7 (rejected) x = y + 6 = 7 Thus the two-digit number is 71 Remarks: 十進制數字 abcd = a*1000 + b*100 + c*10 + d This is what we called "place value"(位值)
Let the 10th digit be x and the unit digit be y. So the original value of the number = 10x + y, when reversed, the value of the number becomes 10y + x, so (10x + y) - (10y + x) = 54 9x - 9y = 54 x - y = 6 or x = 6 + y. also, x^2 + y^2 = 50 that is (6 +y)^2 + y^2 = 50 36 + 12y + 2y^2 - 50 = 0 2y^2 + 12y - 14 = 0 y^2 + 6y - 7 = 0 (y +7)(y-1) = 0 y = -7(rej.) or y = 1 when y = 1, x = 6 + y = 7. so the number is 71.
functions of equations
發問:
a two- digit number is decreased by 54 when its digits are reversed .if the sum of the squares of the digits is 50, find the number
最佳解答:
Let the two digit number be 10x+y then, we have x2 + y2 = 50 ---- (1) and 10y + x = ( 10x + y ) - 54 that is, 9x - 9y = 54 x - y = 6 ---- (2) sub x = y + 6 into (1) gives (y+6)2 + y2 = 50 2y2 + 12y + 36 - 50 = 0 y2 + 6y - 7 = 0 y = 1 or -7 (rejected) x = y + 6 = 7 Thus the two-digit number is 71 Remarks: 十進制數字 abcd = a*1000 + b*100 + c*10 + d This is what we called "place value"(位值)
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其他解答:Let the 10th digit be x and the unit digit be y. So the original value of the number = 10x + y, when reversed, the value of the number becomes 10y + x, so (10x + y) - (10y + x) = 54 9x - 9y = 54 x - y = 6 or x = 6 + y. also, x^2 + y^2 = 50 that is (6 +y)^2 + y^2 = 50 36 + 12y + 2y^2 - 50 = 0 2y^2 + 12y - 14 = 0 y^2 + 6y - 7 = 0 (y +7)(y-1) = 0 y = -7(rej.) or y = 1 when y = 1, x = 6 + y = 7. so the number is 71.
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