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Standard VIII

Mathematics

Taking All the Combinations of 3 Digit Numbers and Adding Them

There is a si...

Question

There is a six digit number with digit $a b c a b c .$
a. The number is multiple of $11$ .
b. One of the factors of the number is $7$ .
c. $1001$ divides the number.

a & c are true

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b & c are true

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None of the above three are true

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All three are true

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Solution

The correct option is D All three are true
$a b c a b c = a b c \times 1000 + a b c$
$= a b c \times (1001) = a b c \times 13 \times 11 \times 7$
$∴$ All the statements are true.

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Similar questions

Q. ABCABC is a six digit number where A,B,C are three consecutive natural numbers. Find the remainder when ABCABC is divided by 1001?

Q. Consider three sets

A, B, C

such that set A contains all three digit numbers that are multiples of

4,

set

B

contains all three digit even numbers that are multiples of

3

and set

C

contains all three digit numbers that are multiples of

5.

Then the number of elements are present in

n (A \cup B \cup C)

Q. Consider three sets

A, B, C

such that set A contains all three digit numbers that are multiples of

4,

set

B

contains all three digit even numbers that are multiples of

3

and set

C

contains all three digit numbers that are multiples of

5.

Then the value of

\frac{n [(A \cap B) \times (A \cap C)]}{n [(B \cap C) \times (A \cap B \cap C)]}

Q. If

a, b,

and

c

are three digits of a three-digit number, prove that

a b c + c a b + b c a

is a multiple of

Q. Consider three sets

A, B, C

such that set A contains all three digit numbers that are multiples of

4,

set

B

contains all three digit even numbers that are multiples of

3

and set

C

contains all three digit numbers that are multiples of

5.

Then the value of

\frac{n [(A \cap B) \times (A \cap C)]}{n [(B \cap C) \times (A \cap B \cap C)]}

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There is a six digit number with digit abcabc.a. The number is multiple of 11.b. One of the factors of the number is 7.c. 1001 divides the number.

The correct option is D All three are trueabcabc=abc×1000+abc=abc×(1001)=abc×13×11×7∴ All the statements are true.

There is a six digit number with digit $a b c a b c .$
a. The number is multiple of $11$ .
b. One of the factors of the number is $7$ .
c. $1001$ divides the number.

The correct option is D All three are true
$a b c a b c = a b c \times 1000 + a b c$
$= a b c \times (1001) = a b c \times 13 \times 11 \times 7$
$∴$ All the statements are true.