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<blockquote data-quote="rajitha_ks" data-source="post: 8016085" data-attributes="member: 162025"><p>wtf dude</p><p>according to ur steps</p><p>lets follow the above steps for the arbitrary number "x" and see</p><p></p><p>1) x</p><p>2) x*3</p><p>3) x*3 + 3 = 3(x+1)</p><p>4)3*3(x+1) = 9(x+1)</p><p></p><p>now you can see that whatever the number you pick you'll end up with a <span style="color: Red"><strong>multiple of 9!</strong></span></p><p></p><p>now these are the multiples of nine</p><p> 9 -> 9+0 = 9</p><p>18 -> 1+8 = 9</p><p>27 -> 2+7 = 9</p><p>36 -> 3+6 = 9</p><p>45 -> 4+5 = 9</p><p>54 -> 5+4 = 9</p><p>63 -> 6+3 = 9</p><p>.</p><p>.</p><p>.</p><p>now if you add the first two digits of all these numbers you'll end up with nine<img src="/styles/default/xenforo/smilies/default/yes.gif" class="smilie" loading="lazy" alt=":yes:" title="Yes :yes:" data-shortname=":yes:" /></p><p></p><p>so nomatter what the number you initially pick, you are bound to end up with 9, which is <span style="color: red"><strong><em><span style="font-family: 'Trebuchet MS'"> Nalith</span></em></strong></span><span style="color: red"></span></p><p><span style="color: red"></span></p></blockquote><p></p>
[QUOTE="rajitha_ks, post: 8016085, member: 162025"] wtf dude according to ur steps lets follow the above steps for the arbitrary number "x" and see 1) x 2) x*3 3) x*3 + 3 = 3(x+1) 4)3*3(x+1) = 9(x+1) now you can see that whatever the number you pick you'll end up with a [COLOR=Red][B]multiple of 9![/B][/COLOR] now these are the multiples of nine 9 -> 9+0 = 9 18 -> 1+8 = 9 27 -> 2+7 = 9 36 -> 3+6 = 9 45 -> 4+5 = 9 54 -> 5+4 = 9 63 -> 6+3 = 9 . . . now if you add the first two digits of all these numbers you'll end up with nine:yes: so nomatter what the number you initially pick, you are bound to end up with 9, which is [COLOR=red][B][I][FONT=Trebuchet MS] Nalith[/FONT][/I][/B][/COLOR][COLOR=red] [/COLOR] [/QUOTE]
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