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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 $$9$$?
    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.
    Since sum of its digits $$ = 8 + 5 + 7 + 6 + 9 + 0 + 1 = 36$$
    $$36$$ is divisible by $$9$$
    Option (a) is the correct answer
  • Question 2
    1 / -0
    The expanded form of the number $$9578$$ is 
    Solution
    Expanded form of $$9578$$
    $$=9\times 1000+5\times 100+7\times 10+8\times 1$$
    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 $$B$$ is right 
    As we know that every number can be arranged as a line.
    The number $$10$$ 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: $$3\times 10000+7\times 1000+9\times 100+0\times 10+4$$
    Solution
    Consider, $$3\times 10000+7\times 1000+9\times 100+0\times 10+4$$
    $$=30000+7000+900+0+4$$
    $$=37904$$
    Hence, option C is right.
  • Question 5
    1 / -0
    Which of the following numbers is divisible by $$ 9 $$ ? 
    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 $$ = 9 + 7 + 2 + 0 + 3 = 27 $$ which is divisible by $$ 9 $$ 
  • Question 6
    1 / -0
    Which of the following numbers is divisible by $$ 3 $$ ?
    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 $$ = 4 + 2 + 0 + 6 + 3 = 15 $$ which is divisible by $$ 3 $$ 
  • Question 7
    1 / -0
    A four-digit number $$aabb$$ is divisible by $$55$$. Then possible value(s) of $$b$$ is/are
    Solution
    If the number is divisible by $$55$$,
    It must also be divisible by $$5$$. 
    Therefore the number ends with $$0$$ or $$5$$.
  • Question 8
    1 / -0
    If $$5A\times A=399$$, then the value of $$A$$ is
    Solution
    Given,
    $$\ \ \ \  5\ A$$
    $$\times\ \ \ \  A$$
    $$\overline{ \ \ 3\ 9\ 9\ \ }$$

    Here,
    $$A\times A=$$ a number whose unit digit is $$9$$
    The number from $$0$$ to $$9$$ whose product with itself is a number having its unit's digit $$9$$ is $$7$$.

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

    Therefore, the six-digit number formed by repeating a three-digit number is always divisible by $$1001$$.
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