Find two consecutive number with a product of 4160

edited

If you put this into an equation it would be as follows

F = First Number

S = Second Number

F*S=4160

Lets try F=50 * S=51

that gives 2550 so it must be higher

F=60 S=61 Gives 3660 so a bit higher

F=64 S=64 Gives 4160. Those are the correct numbers.

F=64

S=65

by Level 1 User (260 points)
Guessing? And guessing again? And guessing a third time? This site is about SOLVING math problems, not guessing.
belThe problem can be stated as follows. Let x be the first number.  The consecutive number is x +1,  The product is x(x+1) = 4160 which can be restated as x^2 +x -4160 =0.  Now sove the roots of the quadratic equation.
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Product of two consequtive numbers:(x)*(x-1)=4160

x^2-x=4160

X^2=4160+x

65^2=4225

4225-65=4160

So two numbers will be:64,65

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First, seek some squared #s nearer to 4160. 60x60=3600, 70x70=4900. Therefore, the consecutive #s are between 60 and 70. Since ones-digit of 4160 is 0, the product of ones-digits of the consecutive two #s should be 0. This indicates consecutive two ones-digits would be 4 and 5, or 5 and 6. 0 and 1, or 9 and 0 also makes the ones-digit 0, but they are omitted because 60x61=60(60+1)=3600+60=3660, 69x70=(70－1)70=4900－70=4830. Therefore, the consecutive #s would be 64 and 65, or 65 and 66. Since 65 is common to both combinations, let try to seek the value of 65x65 using an equation below. (10a+5)x(10a+5)=100xaxa+100xa+5x5=ax(a+1)x100+25. Here a=6, thus 65x65=6x7x100+25=4200+25=4225. 4225 is greater than 4160, and 65x66 is greater than 4225. Therefore, the other # is 64. Let check the product. 64x65=60(10+4+5)+4x5=60x69+4x5=4140+20=4160. 64 and 65 are the correct #s. Since the product of two negative #s are also positive, -64 and -65 are also satisfy the conditions. Therefore, the answers are two set of consecutive #s: {64,65} and {-65,-64}.
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