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Published on: 03/09/2019
Playing with Numbers
Download CBSE Class 6th Standard CBSE Mathematics question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 6th Standard CBSE Mathematics
Questions + Answers key
Take MCQ Mathematics Test

1.
Express the following as the sum of two odd prime numbers 34
2.
Write all the factors of the following number 60
3.
Write all the factors of the following number: 27
4.
State which of the following is correct prime factorisation of numbers.
2x3x3x3x4=216
2x3x7x7x7x11=22638
5.
Which of the following is divisible by both 8 and 12?
1312
1626
1612
1926
None of the given number is divisible by 8 and 12
6.
A factor of 29161137 is
13
7
23
83
7.
A number with 4 or more digits is divisible by 8 if the number formed by the last three digits is divisible by 8.
8.
If a number exactly divides two numbers separately, it must exactly divide their sum.
9.
If a number is divisible by 3, it must be divisible by 9
10.
2 is the only even prime number.
11.
Find the greatest number, which divides 753 and 1054 leaving remainder 3 and 4 respectively.
12.
Express each of the following as the sum of three odd prime numbers. 49
13.
Using divisibility tests, determine the following number is divisible by 6? 1258
14.
Write a digit in the blank space of the following number, so that the number formed is divisible by 11. 8___9484
15.
15 is a multiple of _________ and __________
16.
A number for which the sum of all its factors is equal to twice the number is called a ________________number.
17.
Fill in the given space of prime factorisation of 47957. 7x13x ____x17
1.
Here, we have 34
34 = 31 + 3 or 29 + 5
2.
1, 2, 3, 4, 5, 6,10,12,15,20,30,60
3.
We have, 27
27 = 1x 27; 27 = 3 x 9
\(\therefore\)Factors of 27 are 1, 3, 9 and 27.
4.
(b)
2x3x7x7x7x11=22638
5.
(e)
None of the given number is divisible by 8 and 12
6.
(d)
83
7.
(a)
8.
(a)
9.
(b)
10.
(a)
11.
The required number is the HCF of (753 - 3) and (1054 - 4) i.e. 750 and 1050.

\(\therefore\)The HCF of 750 and 1050 is 150.
Hence, the required number is 150.
12.
49 = 3 + 5 + 41
13.
We have, 1258
(i) Divisibility by 2
\(\because\)Unit digit of number = 8, so 1258 is divisible by 2.
(ii) Divisibility by 3
Sum of digits of given number = 1+ 2 + 5 + 8 = 16 is not divisible by 3, so 1258 is not divisible by 3. Now, we see that 1258 is divisible by 2 but not divisible by 3.
Hence, it is not divisible by 6.
14.
Let the required unknown digit be x.
Then, the number becomes
\(\begin{matrix} 8 \\ \downarrow \\ E \end{matrix}\begin{matrix} x \\ \downarrow \\ O \end{matrix}\begin{matrix} 9 \\ \downarrow \\ E \end{matrix}\begin{matrix} 4 \\ \downarrow \\ O \end{matrix}\begin{matrix} 8 \\ \downarrow \\ E \end{matrix}\begin{matrix} 4 \\ \downarrow \\ O \end{matrix}\)
Sum of digits at odd places from right = 4 + 4 + x = 8+x
Sum of digits at even places from right = 8 + 9 + 8= 25
\(\because\) Number is divisible by 11.
\(\therefore\) Difference of digits will be 0 or 11.
\(\Rightarrow\)25 - (8 + x) = 0 or 11
\(\Rightarrow\)25 - 8 - x = 0 or 11 \(\Rightarrow\) 17 - x = 0 or 11
Taking difference 0, 17-x=0 \(\Rightarrow\) x=17+0 \(\Rightarrow\) x=17 [but 17 is not a single digit number, so it is not possible]
Taking difference 11, 17 - x = 11 \(\Rightarrow\) x = 17 - 11= 6
So, required digit to write in the blank space is 6.
15.
( )
3 and 5
16.
( )
Perfect Number
17.
( )
31
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