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  • #98562

    Problem 5

    A standard dice has number from 1 to 6

    Max product possible= \( 6^2 \) = 36

    Possible combinations are-

    1x6= Sum as 7
    2x3= Sum as 5
    5x6= Sum as 11
    4x3= Sum as 7
    4x6= Sum as 10
    3x6= Sum as 9
    2x6= Sum as 8

    Sum I didn't obtain (Listed in the options)= B(6)

    #98563

    Problem 1 :In the first question, we have_ 222,222-22,222-2,222-222-22-2 we can club each numbers into groups of two and we notice the last digit is always 0 and finally we can conclude that , by subtracting them the end product will also have units digit zero.

    #98564

    Problem 2 : we can use simple calculation methods to simplify 44/11 as 4 , 110/44 as 2.5 and 44/110 as 0.04 and get the result as 6.54

    #98565

    Problem 3: we can find the area of the larger squares ( the outline of gray regions) and from it the area of white square in it . By doing so we get :             (7*7-4*4)+(10*10-9*9) = (49-16)+(100-81)=33+19=52

    #98566

    Problem 4 : first , we find the sum of numbers from 1 to 9 .

    By using the arithmetic progression summation formula we get the sum of first n natural numbers as (n)(n+1)/2.

    By putting 9(as there are 9 terms) in the same , we get 9*10/2=45.

    Now we find the nearest perfect sq. LESS THAN 45 ( coz no sum from 1 to 9  can be greater than which we have taken as no number here is omitted and since our protagonist has missed a number from one to nine his sum is bound to less than 45 ).

    We see that the number satisfying these conditions is 36 . Now we subtract:    45-36 =9 . So He must've missed 9 .

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