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Check whether 2156 is divisible by 7 and 11? Verify whether 2156 is divisible by the product of 11 and 7?

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Last updated date: 17th Jun 2024
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Answer
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Hint: We find out the concept of multiplier. If the given numbers are formed as a multiplication form of two positive whole numbers then we can assure about the divisibility. We also check whether it is directly divisible by 7 or 11 using the division formulas.

Complete step-by-step solution:
The concept of a number being divisible by another number comes from the concept of multiplicity which is the expression of the first number being termed as a multiplication form of two other positive whole numbers different from the given first number.
For the divisibility of 7, we know that we take twice the unit value. Then we subtract that value from the remainder of the original number. If that subtracted value is divisible by 7 then the number itself is divisible by 7.
For example, we take a number (mno) where m, n, o are the digits in that number in the hundredth, tenth, unit places. So, we find o. take the twice of the digit which is 2o. The remainder of the number excluding the unit digit is (mn). We subtract these two and if that value is divisible by 7 then we can say that (mno) is divisible by 7. Take 2689. We take unit digit and get twice of it which is $9\times 2=18$. Now, we take subtraction of 18 from 268, the remaining of the number. We get $268-18=250$ which is not divisible by 7. So, 2689 is not divisible by 7 where $\dfrac{2689}{7}=384.14$. If the number is very long then we can continue the same process as many times as we want.
For the divisibility of 11, we know that if the subtraction of the sum value of the oddly placed digits from the even placed digits is divisible by 11 then the number itself is divisible by 11.
For example, we take a number (abc) where a, b, c are the digits in that number in the hundredth, tenth, unit places. So, we find odd placed digits which are a and c. the even placed digits are b. the subtraction value is $\left( a+c \right)-b$. If this divisible by 11 then (abc) is divisible by 11. Let’s take in 3654. Here $\left( 6+4 \right)-\left( 3+5 \right)=2$ which is not divisible by 11. So, 3645 is not divisible by 11 where $\dfrac{3645}{11}=331.36$.
Now we talk about 2156. It has 6 in unit place. The twice value is 12. So, the difference is $215-12=203$. It’s divisible by 7 as $\dfrac{203}{7}=29$. So, 2156 is divisible by 7 where $\dfrac{2156}{7}=308$.
2156 has 2 and 5 in even-numbered place. It also has 1 and 6 in odd-numbered place. The difference is $\left( 5+2 \right)-\left( 6+1 \right)=0$ which is divisible by 11. So, 2156 is also divisible by 11. Where $\dfrac{2156}{11}=196$.
So, 2156 is divisible by both 7 and 11.
Now we check if the product of 7 and 11 divides 2156 or not.
So, the form is $\dfrac{2156}{7\times 11}$. We got that 2156 is divisible by 11. So, if the result is divisible by 7 then we can say that the product of 7 and 11 divides 2156.
We got $\dfrac{2156}{11}=196$. We need to check whether 196 is divisible by 7 or not.
Now, applying the theorem we got the subtracted value as $19-\left( 6\times 2 \right)=7$ which is divisible by 7. So, 196 is also divisible by 7 where $\dfrac{196}{7}=28$.
So, 2156 is divisible by the product of 11 and 7.

Note: We need to remember the theorems for the divisibility. If we can’t then there will be no way to find out other than doing long division. Using the theorem saves a lot of time. The double divisibility or the divisibility of the product form can also be calculated by first getting the value of the product. The multiplication form is also very important to establish the statement as true.