 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Square Your Age (Posted on 2003-04-15) The other day, Jim excitedly told me, "Did you realise I will turn x years old in the year x^2?"

He wasn't the first to think of this. The 19th century mathematician August de Morgan used to used to boast that he was x years old in the year x^2. He died in 1871.

In what year was Jim born? When will his prediction be true? In what year was de Morgan born? What is x in each case?

 See The Solution Submitted by pleasance Rating: 2.7500 (8 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Alternative Methodology Comment 20 of 20 | (In reply to Solution To The Problem by K Sengupta)

De Morgan's Age

As before in the earlier post, we will get the follwing inequation:

1800< x^2 - x< 1900 ......(#)

Now, x^2 -x > 1800. Since 42^2 - 42 = 1722 < 1800 < 1806 = 43^2, we must have x >=43.

Again, x^2 -x < 1900.

Now,  43^2 - 43 = 1806 < 1900 > 1892 = 44^2 - 44, which is a contradiction.

Again, 44^2 - 44 = 1892 < 1900 < 1980 = 45^2 - 45, and therefore x<=45

Since 43<= x <= 45, substituting x = 43, 44, 45, in (#), we observe that only x = 43 satisfies the referred to inequality.

Consequently, de Morgan was born in 1806  and he was precisely 1849 - 1806 = 43 years old in the year 1849 (43^2)

Jim's Prediction

Considering the inequality 1900< x^2 - x< 2000, and proceeding in a similar manner, we observe that the only positive integer x for which this inequality is satisfied ocurs at x = 45.

Jim was born in the year 1980 and the prediction will come true in the year 2025.

 Posted by K Sengupta on 2009-01-30 02:23:48 Please log in:
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