Advertisements
Advertisements
Question
The sum of all the numbers formed by the digits x, y and z of the number xyz is divisible by ______.
Options
11
33
37
74
MCQ
Fill in the Blanks
Solution
The sum of all the numbers formed by the digits x, y and z of the number xyz is divisible by 37.
Explanation:
We have, xyz + yzx + zxy
= (100x + 10y + z) + (100y + 10z + x) + (100z + 10x + y) ...(i)
= 100x + 10x + x + 10y + 100y + y + z + 100z + 10z
= 111x + 111y + 111z
= 111(x + y + z)
= 3 × 37 × (x + y + z)
Hence, equation (i) is divisible by 37, but not divisible by 11, 33 and 74.
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
Solve each of the following Cryptarithm:
A B
+3 7
9 A
Solve each of the following Cryptarithm:
2 A B
+ A B 1
B 1 8
Solve each of the following Cryptarithm:
1 2 A
+ 6 A B
A 0 9
Show that the Cryptarithm
\[4 \times \overline{{AB}} = \overline{{CAB}}\] does not have any solution.
Write the following number in generalised form.
73
Write the following number in generalised form.
129
Write the following in the usual form.
100 × 7 + 10 × 1 + 8
Write the following in the usual form:
100 × a + 10 × c + b
The usual form of 1000a + 10b + c is ______.
If 5A + B3 = 65, then the value of A and B is ______.
