Live now! Full branding package giveaway worth $11,895. Two Interesting Math Problems Problem1: Smallest Autobiographical Number: A number with ten digits or less is called autobiographical if its first digit (from the left) indicates the number of zeros it contains,the second digit the number of ones, third digit number of twos and so on. For example: 42101000 is autobiographical. Find, with explanation, the smallest autobiographical number. Solution of Problem 1 Problem 2: Fit Rectangle: A rectangle has dimensions$ 39.375$cm$ \times 136.5$cm. • Find the least number of squares that will fill the rectangle. • Find the least number of squares that will fill the rectangle, if every square must be the same size and Find the largest square that can be tiled to completely fill the rectangle. Solution of Problem 2 Solutions of Problem 1: The restrictions which define an autobiographical number make it straightforward to find the lowest one. It cannot be 0, since by definition the first digit must indicate the number of zeros in the number. Presumably then, the smallest possible autobiographical number will contain only one 0.If this is the case, then the first digit must be 1. 10 is not a candidate because the second digit must indicate the number of 1s in the number–in this case, 1. So If the number contains only one zero, it must contain more than one 1. (If it contained one 1 and one 0, then the first two digits would be 11, which would be contradictory since it actually contains two 1s). Again, presumably the lowest possible such number will contain the lowest possible number of 1s, so we try a number with one 0 and two 1s. It will be of the form: 12-0–.. Now, there is one 2 in this number, so the first three digits must be 121. To meet all the conditions discussed above, we can simply take a 0 onto the end of this to obtain 1210, which is the smallest auto-biographical number. Solution of Problem 2: We solve the second and third parts of the question first: We convert each number to a fraction and get a common denominator, then find the gcd (greatest common divisor) of the numerators. That is, with side lengths$ 39.375$cm and$ 136.5$cm , we convert those numbers to fractions (with a common denominator):$ 39.375 = \dfrac{315}{8}$.$ 136.5 = \dfrac{273}{2} = \dfrac{1092}{8}$. Now we need to find the largest common factor of 315 and 1092. Which is 21. So$ \dfrac{21}{8}=2.625$is the largest number that divides evenly into the two numbers$ 39.375$and$ 136.5$. There will be$ \dfrac{1092}{21} \times \dfrac{315}{21} = 52 \times 15 = 780$squares, each one a$ 2.625$cm$ \times 2.625\$ cm square needed to fill the rectangle (52 in each row,with 15 rows).

Now we shall solve the first part.
Number of squares lengthwise is 52 and breadthwise is 15. Now we will combine these squares in order to find least number of squares to fill the rectangle. First three squares would be of
dimension 15 by 15. In this way length of 45 units is utilized. Now the rectangle which is left with us excluding three squares is 7 by 15. Again in the same way we can make two squares of dimension 7 by 7. In this way breadth of 14 units is utilized.
Now we are left with the rectangle of dimension 7 by 1.
These can further be subdivided into seven squares each of
dimension 1 by 1. In this way the least number of squares to fill the
rectangle is 3 + 2+ 7 = 12. The required answer is 12.
Note that the three numbers 3, 2, and 7 are involved in the Euclidean Algorithm for finding the g.c.d.!

Source: Internet

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