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Playing with Numbers Test - 18

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Playing with Numbers Test - 18
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Which of the following numbers is divisible by 99?
    Solution
    We use a rule to check whether the number is divisible by 9 or not.
    A number is divisible by 9 if the sum of the digits is evenly divisible by 9.
    Since sum of its digits =8+5+7+6+9+0+1=36 = 8 + 5 + 7 + 6 + 9 + 0 + 1 = 36
    3636 is divisible by 99
    Option (a) is the correct answer
  • Question 2
    1 / -0
    The expanded form of the number 95789578 is 
    Solution
    Expanded form of 95789578
    =9×1000+5×100+7×10+8×1=9\times 1000+5\times 100+7\times 10+8\times 1
    Hence, B is the correct answer.
  • Question 3
    1 / -0
    By using dot (.)(.) patterns, which of the following numbers can be arranged in all the three ways namely a line, a triangle and a rectangle ?
    Solution
    Option BB is right 
    As we know that every number can be arranged as a line.
    The number 1010 can be shown as 
    ....................
    Also, it can show a triangle Given below
    And, the number cam show as a rectangle. Given below

  • Question 4
    1 / -0
    Evaluate: 3×10000+7×1000+9×100+0×10+43\times 10000+7\times 1000+9\times 100+0\times 10+4
    Solution
    Consider, 3×10000+7×1000+9×100+0×10+43\times 10000+7\times 1000+9\times 100+0\times 10+4
    =30000+7000+900+0+4=30000+7000+900+0+4
    =37904=37904
    Hence, option C is right.
  • Question 5
    1 / -0
    Which of the following numbers is divisible by 9 9
    Solution
    We use a rule to check whether the number is divisible by 9 or not.
    A number is divisible by 9 if the sum of the digits is evenly divisible by 9.
    Sum of digits =9+7+2+0+3=27 = 9 + 7 + 2 + 0 + 3 = 27 which is divisible by 9 9  
  • Question 6
    1 / -0
    Which of the following numbers is divisible by 3 3 ?
    Solution
    We use a rule to check whether the number is divisible by 3 or not.
    A number is divisible by 3 if the sum of the digits is evenly divisible by 3.
    Sum of digits is =4+2+0+6+3=15 = 4 + 2 + 0 + 6 + 3 = 15 which is divisible by 3 3  
  • Question 7
    1 / -0
    A four-digit number aabbaabb is divisible by 5555. Then possible value(s) of bb is/are
    Solution
    If the number is divisible by 5555,
    It must also be divisible by 55
    Therefore the number ends with 00 or 55.
  • Question 8
    1 / -0
    If 5A×A=3995A\times A=399, then the value of AA is
    Solution
    Given,
    $$\ \ \ \  5\ A$$
    $$\times\ \ \ \  A$$
      3 9 9  \overline{ \ \ 3\ 9\ 9\ \ }

    Here,
    A×A=A\times A= a number whose unit digit is 99
    The number from 00 to 99 whose product with itself is a number having its unit's digit 99 is 77.

    Hence, A=7A=7
    And the option (C) is correct.
  • Question 9
    1 / -0
    Let abcabc be a three digit number. Then abc+bca+cababc+bca+cab is not divisible by
    Solution
    We have,
    abc+bca+cab=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)abc+bca+cab=(100a+10b+c)+(100b+10c+a)+(100c+10a+b)
    =111a+111b+111c=111a+111b+111c
    =111(a+b+c)=111(a+b+c)
    =3×37(a+b+c)=3\times 37(a+b+c)
    Therefore the number abc+bca+cababc+bca+cab is divisible by3,373,37 and (a+b+c)(a+b+c)  but not by 99.
  • Question 10
    1 / -0
    A six-digit number is formed by repeating a three-digit number. For example 256256,678678256256, 678678, etc. Any number of this form is always divisible by
    Solution
    Let
    A six-digit number formed by repeating a three-digit number be abcabcabcabc
    Then,
    abcabc=100000a+10000b+1000c+100a+10b+cabcabc=100000a+10000b+1000c+100a+10b+c
    =a(100000+100)+b(10000+10)+c(1000+1)=a(100000+100)+b(10000+10)+c(1000+1)
    =(100100)a+(10010)b+(1001)c=(100100)a+(10010)b+(1001)c
    =1001(100a+10b+c)=1001(100a+10b+c)

    Therefore, the six-digit number formed by repeating a three-digit number is always divisible by 10011001.
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