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Write the smallest 7 digit number ending in 5.

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Hint: Here in this question, we have to write the smallest number where the unit place will be 5and the number should be 7 digits. We should know about the numbers and by knowing the place value system and we write the number. By using the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

Complete step-by-step answer:
In mathematics we have different kinds of numbers namely, natural numbers, whole numbers, integers, rational numbers, irrational numbers and real numbers.
Natural numbers are defined as counting numbers.
Whole numbers are defined as the counting numbers and along with it 0.
Integers are defined as positive and negative numbers of whole numbers.
Rational numbers are defined as the numbers is in the form of \[\dfrac{p}{q}\]
Irrational numbers are defined as the numbers which are not rational.
Real numbers are defined as the combination of integers and rational numbers.
The place value system for 7 digits is defined by

TenLakhLakhTenThousandThousandHundredtenones
10,00,0001,00,00010,0001,000100101


Consider the Whole numbers from 0 to 9
The even numbers are 0, 2, 4, 6, 8
The odd numbers are 1, 3, 5, 7, 9
here we have to form the smallest 7 digit numbers and the unit place is fixed and that is 5. The least number from 0 to 9 is 0, therefore from lakh to tens place we will write the number 0. In the ten lakh place we will write the number 1. Suppose if we write 0 in the ten lakh place then it will not become a 7 digit number.
Therefore the smallest 7 digit even number is 10,00,005.
So, the correct answer is “10,00,005”.

Note: In mathematics we have different kinds of numbers, here while writing the number we choose the numbers which belong to the whole numbers, here we can’t take the numbers from the natural numbers because the natural numbers does not include 0. While writing the digits we need the number 0.