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Note that 7 | 42 and 7 | 77, meaning 42 and 77 are multiples of 7. Through addition of multiples of these integers, we can identify other multiples of 7. Fill in the blanks with whole numbers to create true statements of arithmetic in which all of the sums are multiples of 7.  

____ x 42 + ____ x 77 = 350 = 50 x 7 

 Jun 10, 2023
 #1
avatar+754 
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Note that 7 | 42 and 7 | 77, meaning 42 and 77 are multiples of 7. Through addition of multiples of these integers, we can identify other multiples of 7. Fill in the blanks with whole numbers to create true statements of arithmetic in which all of the sums are multiples of 7.  

____ x 42 + ____ x 77 = 350 = 50 x 7   

 

Some parentheses would have been helpful, so, lacking them,  

I am going to assume PEMDAS, which would put them thus:  

 

(___ x 42) + (___ x 77)  and I'll assign unknowns to the blanks.  

 

(A x 42) + (B x 77)  =  350  

 

A has to be less than 9 or (A x 42) would be more than 350 by itself.  

B has to be less than 5 or (B x 77) would be more than 350 by itself.  

 

If B is 1, then (A x 42) has to end with 3.  There's no such integer.  

If B is 2, then (A x 42) has to end with 6.  A might be 3 or 8.  

If B is 3, then (A x 42) has to end with 9.  There's no such integer.  

If B is 4, then (A x 42) has to end with 2.  A might be 1 or 6.  

 

(3 x 42) + (2 x 77) = 126 + 154 = 280 ... eliminate because it doesn't total 305.  

(8 x 42) + (2 x 77) = 336 + 154 = 490 ... eliminate because it doesn't total 350.

(1 x 42) + (4 x 77) =   42 + 308 = 350 ... a possible, but let's try the last one.  

(6 x 42) + (4 x 77) = 252 + 308 = 560 ... eliminate because it doesn't total 305.  

 

So the only numbers that will work are (1 x 42) + (4 x 77) = 350  

 

 Jun 11, 2023
 #2
avatar+754 
+2

 

oops, on the next to last line I rolled a couple of digits.  

 

I hope it was obvious that the "305" should be "350".  

 

Bosco  Jun 12, 2023

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