Find the minimum value of a + b + c + ⋯ + z when
a × b × c × ⋯ × z = 1 0 0 .
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Nice problem. Next time whenever you share any problem, don't forget to level it. @Pratik Shastri
Le problème est un peu flou.
If all letters except for z are very large negative numbers, then z has to be a very small negative number in order for all letters to multiply to 100. (E.g. consider − b where b is very large). This means that the minimum value consists of 25 negative big numbers and one negative small number. This adds to a very big negative number. Hence, the larger all letters (but negative and large such as − 1 0 0 0 0 0 0 0 0 0 0 except for z , the smaller the minimum and hence it tends to − ∞ as you can make all letters except for z negatively larger and larger.
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I presume that we are working in real number system ! So Minimum can be − ∞ . because it is not necessary that a,b,c,..........z are positive real's . (So option 2 6 2 6 1 0 0 . is rejected )
without loss of generality Let a = lim a → − ∞ ( a ) . and also Let b = lim b → 0 ( b ) . and again let al other alphabhet's (Real number's ) are aribitary number's which has finite representation in Real number system ( say 2 ,-8 , +11 , -19, 2 , 3 1 7 2 9 . ....etc.)
So by fundamental rule of limit lim a → − ∞ ( a ) × lim b → 0 ( b ) = L ( c a n b e e x i s t ) .
And According to question It must has to exist (since a ∗ b ∗ c ∗ d . . . . . . . . z = 1 0 0 . )
Therefore ( a + b + c + . . . . + z ) m i n ⟶ − ∞ .