Question

what is the maximum value of q, 5p7+8q9+r32=1928

Answer 1

Answer:

8

Step-by-step explanation:

5p7

+8q9

+r32

=19(p+q+4)8

:, p+q+4=12

p+q=8Therefore max value of q=8

Hence proved,,.

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Answer 2

Answer:

First find the sum of unit digits,

7+9+2=18

1 is forward carry

Now sum of tens digit

(P+Q+3)+1=10\dotX+2

P+Q+4=10\dotX+2

Here, X is carry forward

Now the sum of hundreds digit

(5+8+R)+X=19

⟹X=6−R

Now, we have

P+Q+4=10(6−R)+2

⟹P+Q+10R=58

Also P and Q cannot be greater than 9 (why?)

On putting P=Q=9

We get R=4

Therefore maximum value of Q is 9

Check 597+899+432=1928